Practice topic-wise Thermodynamics previous year questions for ISRO Mechanical & Aerospace with instant step-by-step solutions and formula breakdowns.
All ISRO Mechanical & Aerospace papers
62 questions.
Year: 2008 | Topic: Thermodynamics
The density of air at 10 °C and 1MPa abs, in SI units is
Answer: B. 12.31
Explanation: T = 10 °C = 283 K. P = 1 MPa = 10^6 Pa. Using ideal gas equation P = rho * R * T, rho = P / (RT) = 10^6 / (287 * 283) = 12.31 kg/m³.
Year: 2008 | Topic: Thermodynamics
(dp/dV)_T * (dv/dT)_P * (dT/dp)_v is equal to
Answer: C. -1
Explanation: For an ideal gas Pv = RT. (dP/dv)_T = -RT/v². (dv/dT)_p = R/P. (dT/dP)_v = v/R. Multiplying them all: (-RT/v²) * (R/P) * (v/R) = -RT / (Pv) = -1. This is also known as Euler's chain rule.
Year: 2008 | Topic: Thermodynamics
A well insulated vessel containing air is heated to double the pressure and temperature 150% in excess of initial. The amount of air that leaked is
Answer: C. 25%
Explanation: For an ideal gas P1*V1/T1 = P2*V2/T2. Given P2 = 2*P1 and T2 = 1.5*T1. P1*V1/T1 = 2*P1*V2 / 1.5*T1. This gives V1 = (4/3)V2. The amount of air leaked is (V1 - V2)/V1 % = (V1 - 0.75V1)/V1 = 25%.
Year: 2008 | Topic: Thermodynamics
The specific heat of diatomic gas at constant pressure is given by (R = universal gas constant and M = molecular weight of gas)
Answer: C. 7/2 R/M
Explanation: We know that cp = gamma * R_g / (gamma - 1). For diatomic gas, gamma = 1.4. R_g = R/M. cp = 1.4 * (R/M) / (1.4 - 1) = 1.4/0.4 * (R/M) = 7/2 (R/M).
Year: 2008 | Topic: Thermodynamics
Kinetic energy of the molecules in terms of absolute temperature (T) is proportional to
Answer: A. T
Explanation: Kinetic energy of molecules is given by E = (3/2)kT, where k is the Boltzmann constant and T is Absolute temperature. Thus, E is directly proportional to T.
Year: 2008 | Topic: Thermodynamics
The quantity of heat required to change the temperature of 1 gm of ice cream -6 °C to -5 °C is known as
Answer: D. Specific heat
Explanation: Specific heat is defined as the amount of thermal energy needed to raise the temperature of 1 gram of a substance by 1 degree Celsius. Therefore, raising ice cream from -6 °C to -5 °C represents its specific heat.
Year: 2014 | Topic: Thermodynamics
A steam engine intakes 100 gm of steam at 100 C per minute and cools it down to 20 C. Calculate the heat rejected by the steam engine per minute (consider Latent Heat of vaporization of steam = 540 cal/gm and Cp = 1 cal/gm C)
Answer: C. 6.2 x 10^4 cal
Explanation: The total heat rejected per minute involves condensing the steam and then cooling the water. Heat rejected Q_R = (m * Cp * delta_T) + (m * L). Substituting the given values: Q_R = (100 * 1 * (100 - 20)) + (100 * 540) = (100 * 80) + 54000 = 8000 + 54000 = 62000 cal, which is equal to 6.2 x 10^4 cal.
Year: 2014 | Topic: Thermodynamics
A heat engine operates between a cold reservoir at temperature T2 = 300K and a hot reservoir at Temperature T1. It takes 200J of heat from the hot reservoir and deliver 120J of heat to the cold reservoir in a cycle. What will be the efficiency of the engine?
Answer: B. 40%
Explanation: The efficiency of a heat engine is given by the formula eta = W / Q_S = (Q_S - Q_R) / Q_S = 1 - (Q_R / Q_S). Given the heat supplied Q_S = 200 J and the heat rejected Q_R = 120 J, the efficiency is 1 - (120 / 200) = 1 - 0.6 = 0.4, which corresponds to 40%.
Year: 2014 | Topic: Thermodynamics
As shown in Fig. below, an ideal gas is confined to a cylinder by a piston. The piston is pushed down slowly so that the gas temperature remains at 27°C During the compression, 900 J of work is done on the gas. Find the entropy change of the gas.
Answer: B. -3 J/K
Explanation: Applying the first law of thermodynamics, $\partial Q = dU + \partial W$. Since the temperature is kept constant (isothermal compression), the change in internal energy $dU = 0$. Since 900 J of work is done *on* the gas, $\partial W = -900$ J. Thus, heat transfer $\partial Q = -900$ J. The change in entropy is $dS = \frac{\partial Q}{T}$. With $T = 27^{\circ}C = 300$ K, $dS = -\frac{900}{300} = -3$ J/K.
Year: 2014 | Topic: Thermodynamics
A sample of gas ($\gamma=1.5$) is taken through an adiabatic process in which the volume is compressed from $1600~cm^{3}$ to $400~cm^{3}$. If initial pressure is 150 kPa, then how much will be the final pressure in the process?
Answer: A. 1200 kPa
Explanation: For an adiabatic process, the governing equation is $P_{1}V_{1}^{\gamma} = P_{2}V_{2}^{\gamma}$. Rearranging to solve for the final pressure gives $P_{2} = P_{1}(\frac{V_{1}}{V_{2}})^{\gamma}$. Substituting the given values: $P_{2} = 150 \times (\frac{1600}{400})^{1.5} = 150 \times 4^{1.5}$. Since $4^{1.5} = 8$, the final pressure is $P_{2} = 150 \times 8 = 1200\text{ kPa}$.
Year: 2014 | Topic: Thermodynamics
One mole of a monatomic gas is taken through a cycle ABCDA as shown in the P-V diagram. Column II give the characteristics involved in the cycle. Match A $\rightarrow$ B process with the processes given in Column II.
Process A $\rightarrow$ B goes from (3V, 3P) to (1V, 3P) at constant pressure.
Column II: (p) Internal energy decreases, (q) Internal energy increases, (r) Heat is lost, (s) Heat is gained, (t) Work is done on the gas.
Answer: C. p, r, t
Explanation: Process A $\rightarrow$ B occurs at a constant pressure $3P$. According to the ideal gas law at constant pressure, $\frac{V_{A}}{T_{A}} = \frac{V_{B}}{T_{B}}$. Since $V_{B} = 1V$ and $V_{A} = 3V$, we have $T_{B} = \frac{T_{A}}{3}$. Because $T_{B} < T_{A}$, the gas cools down, meaning heat is 'lost' (r). The change in internal energy is $\Delta U = C_{V}(T_{B}-T_{A})$, which is negative, so internal energy decreases (p). Finally, since the volume is compressed from $3V$ to $1V$, work is done on the gas (t). Therefore, the correct characteristics are p, r, and t.
Year: 2015 | Topic: Thermodynamics
The difference between two specific heats, $C_p - C_v = \frac{R}{J}$. This relation is valid for
Answer: B. Perfect gases
Explanation: Mayer's relation $C_p - C_v = R$ (or $R/J$ in appropriate units) strictly applies to ideal or perfect gases. For perfect gases, specific heats ($C_p$ and $C_v$) are assumed not to change with temperature.
Year: 2015 | Topic: Thermodynamics
A process in which no heat is supplied or rejected from the system and entropy is not constant is known as
Answer: C. Polytropic
Explanation: According to the provided solution, if no heat is supplied or rejected and entropy is constant, the process is isentropic. However, if no heat is supplied or rejected and entropy is *not* constant, the provided reference explicitly classifies this as a polytropic process.
Year: 2015 | Topic: Thermodynamics
Carnot cycle has maximum efficiency for
Answer: C. Reversible engine
Explanation: According to Carnot's theorem, no engine operating between two given temperature limits can be more efficient than a completely reversible engine operating between the same limits. Thus, the Carnot cycle has maximum efficiency for a reversible engine.
Year: 2015 | Topic: Thermodynamics
The inlet and outlet conditions of steam for an adiabatic steam turbine are as indicated in the figure. The notations are as usually followed. If the mass rate of steam through the turbine is 20 kg/s, the power output of the turbine is (Inlet: $h_1=3200$ kJ/kg, $V_1=160$ m/s, $Z_1=10$ m; Outlet: $h_2=2600$ kJ/kg, $V_2=100$ m/s, $Z_2=6$ m)
Answer: A. 12.157 MW
Explanation: Applying the Steady Flow Energy Equation (SFEE) for an adiabatic process ($Q = 0$): $m(h_1 + V_1^2/2000 + gZ_1/1000) = m(h_2 + V_2^2/2000 + gZ_2/1000) + W$. Substituting the given values yields $W = 20 \times [(3200 - 2600) + (160^2 - 100^2)/2000 + 9.81 \times (10 - 6)/1000]$. Evaluating this gives $W = 20 \times [600 + 7.8 + 0.03924] = 20 \times 607.83924 = 12156.78$ kW, which is approximately 12.157 MW.
Year: 2015 | Topic: Thermodynamics
A solar energy based heat engine which receives 80 kJ of heat at $100^{\circ}C$ and rejects 70 kJ of heat to the ambient at $30^{\circ}C$ is to be designed. The thermal efficiency of the heat engine is
Answer: C. 12.5%
Explanation: The heat supplied to the engine is $Q_S = 80$ kJ and the heat rejected is $Q_R = 70$ kJ. The net work output is $W_{net} = Q_S - Q_R = 80 - 70 = 10$ kJ. The actual thermal efficiency is defined as $\eta_{th} = W_{net} / Q_S = 10 / 80 = 0.125$, which is $12.5\%$.
Year: 2015 | Topic: Thermodynamics
A gas having a negative Joule-Thomson effect (mu < 0) when throttled will
Answer: B. Become warmer
Explanation: The Joule-Thomson coefficient (mu) indicates the change in temperature with respect to pressure during throttling. In the heating region, mu is negative, meaning that a drop in pressure results in an increase in temperature, thus making the gas warmer.
Year: 2015 | Topic: Thermodynamics
The pressure 'p' of an ideal gas and its mean kinetic energy E per unit volume are related by the relation
Answer: D. p = (2E)/3
Explanation: The average kinetic energy (E) per unit volume of an ideal gas is theoretically given by E = (3/2) * p, where p is the pressure. Rearranging this relationship to solve for pressure yields p = (2/3) * E.
Year: 2015 | Topic: Thermodynamics
Wet clothes are hung on a clothesline outdoors in sub-zero weather. After a day, the clothes are brought into the house and observed to be dry. The process of drying is best explained as
Answer: B. Sublimation
Explanation: In sub-zero temperatures, the water in the wet clothes first freezes into ice. Over time, this ice transitions directly from the solid phase into water vapor (gas) without passing through the liquid phase. This specific phase change is known as sublimation.
Year: 2015 | Topic: Thermodynamics
A 10 kg solid at 100 C with a specific heat of 0.8 kJ/kg C is immersed in 40 kg of 20 C liquid with a specific heat of 4.0 kJ/kg C. Estimate the temperature after a long time if the container is insulated (Specific heat of water 4.18 kJ/kg).
Answer: D. 24 C
Explanation: Using the principle of conservation of energy, the heat given by the solid equals the heat taken by the liquid: 10 * 0.8 * (100 - t) = 40 * 4.0 * (t - 20). Simplifying this gives 800 - 8t = 160t - 3200. Solving for t yields 168t = 4000, so t = 23.8 C, which rounds closely to 24 C.
Year: 2015 | Topic: Thermodynamics
A steel ball of mass 1 kg and specific heat 0.4 kJ/kg is at a temperature of 60 C. It is dropped into 1 kg water at 20 C. The final steady state temperature of water is
Answer: A. 23.5 C
Explanation: The heat lost by the steel ball equates to the heat gained by the water: [m * c * delta_t]_ball = [m * c * delta_t]_water. Substituting the values gives 1 * 0.4 * (60 - t) = 1 * 4.18 * (t - 20). This expands to 24 - 0.4t = 4.18t - 83.6. Rearranging terms yields 107.6 = 4.58t, resulting in t = 23.49 C, which is approximately 23.5 C.
Year: 2016 | Topic: Thermodynamics
Starting from the same initial conditions, an ideal gas expands from volume V1 to V2 in three different ways, the work done by the gas is W1 if the process is purely isothermal, W2 if purely isobaric and W3 is purely adiabatic, then
Answer: A. W2 > W1 > W3
Explanation: The area under the P-V diagram equals the work done during any thermodynamic process. For an expansion from volume V1 to V2, the constant pressure (isobaric) process exhibits the highest area under the curve, followed by the constant temperature (isothermal) process, while the adiabatic process has the least area. Therefore, the work done follows the order W2 > W1 > W3.
Year: 2016 | Topic: Thermodynamics
A house refrigerator with its door open is switched on in a closed room. The air in the room is
Answer: C. heated
Explanation: When the refrigerator door is left open, it absorbs heat (Q2) from the room's space. Simultaneously, heat (Q1) is rejected back into the room from the condenser. Since the compressor continuously does work (W) to run the cycle, the total heat rejected is Q1 = Q2 + W. Because Q1 is strictly greater than Q2 due to the compressor work, there is a net heat addition to the room, causing its temperature to rise.
Year: 2016 | Topic: Thermodynamics
A 10 kW drilling machine is used to drill a bore in a small aluminium block of mass 8 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50% of power is used up in heating the block? (specific heat of aluminum: 0.91 J/(g °C))
Answer: C. 103 °C
Explanation: The total power of the drilling machine is 10 kW, but only 50% is used for heating, which equals 5 kW (or 5000 W). The total heat supplied over 2.5 minutes is Q = Power * Time = 5000 W * (2.5 * 60) s = 750,000 J. Using the heat capacity formula Q = m * C * delta_T, we substitute the mass m = 8 kg and specific heat C = 0.91 * 1000 J/(kg °C) = 910 J/(kg °C). This gives 750,000 = 8 * 910 * delta_T. Solving for the temperature change yields delta_T = 750,000 / 7280 ≈ 103.02 °C.
Year: 2016 | Topic: Thermodynamics
A block of ice at $-10^{\circ}C$ is slowly heated and converted to steam at $100^{\circ}C$. Which of the following curves represent the phenomena qualitatively?
Answer: A.
Explanation: The heating process occurs in stages: (I) Heating of ice from $-10^{\circ}C$ to $0^{\circ}C$ where the temperature increases. (II) Phase change from ice to water at $0^{\circ}C$ where the temperature remains constant. (III) Heating of water from $0^{\circ}C$ to $100^{\circ}C$ where the temperature increases again. (IV) Phase change from water to steam at $100^{\circ}C$ where the temperature remains constant. Curve (a) correctly depicts these alternating temperature rises and constant temperature plateaus.
Year: 2016 | Topic: Thermodynamics
A thermometer works on the principle of
Answer: B. Zeroth law of thermodynamics
Explanation: A thermometer operates on the principle of the Zeroth Law of Thermodynamics. This law states that if two bodies A and B are each in thermal equilibrium with a third body C (such as a thermometer), then A and B are in thermal equilibrium with each other. This allows the thermometer to measure the temperature of a body by coming into thermal equilibrium with it.
Year: 2016 | Topic: Thermodynamics
A mercury thermometer was first placed in melting ice and the length of mercury column was observed to be 10 mm; when it was placed in steam, the length of the column was 250 mm. When placed in tap water, the length of the column was 58 mm. The temperature of the tap water is
Answer: B. 20°C
Explanation: Assuming a linear temperature profile, the equation is $T = al + b$. For melting ice ($0^{\circ}C$, $l = 10 mm$), $0 = 10a + b$. For steam ($100^{\circ}C$, $l = 250 mm$), $100 = 250a + b$. Subtracting the first equation from the second gives $100 = 240a$, leading to $a = \frac{5}{12}$. Substituting $a$ back into the first equation yields $b = -10 \times \frac{5}{12} = -\frac{25}{6}$. The temperature equation is $T = \frac{5}{12}l - \frac{25}{6}$. For tap water ($l = 58 mm$), $T = \frac{5}{12}(58) - \frac{25}{6} = 20^{\circ}C$.
Year: 2016 | Topic: Thermodynamics
The amount of steam (at $100^{\circ}C$) required to raise the temperature of 200 g of water from $60^{\circ}C$ to $100^{\circ}C$ is
Answer: D. 14.8 g
Explanation: The heat gained by the water must equal the heat lost by the condensing steam. Heat gained by water = $m_w \times c_p \times \Delta T = 0.2 kg \times 4.187 kJ/(kg\cdot K) \times 40 K$. The heat released by the steam during phase change is $m_s \times L$, where $L$ is the latent heat of steam ($2257 kJ/kg$). Equating these: $m_s \times 2257 = 0.2 \times 4.187 \times 40$. Solving for $m_s$ provides 14.8 g.
Year: 2016 | Topic: Thermodynamics
A engine works on Carnot cycle between $727^{\circ}C$ and $227^{\circ}C$. The efficiency of the engine is
Answer: A. 50%
Explanation: Convert the given temperatures to Kelvin: the lower temperature $T_L = 227 + 273 = 500 K$, and the higher temperature $T_H = 727 + 273 = 1000 K$. The efficiency of a Carnot engine is calculated by $\eta = 1 - \frac{T_L}{T_H} = 1 - \frac{500}{1000} = 0.5$, which corresponds to 50%.
Year: 2016 | Topic: Thermodynamics
What parameter will remain constant in a throttling process?
Answer: D. Enthalpy
Explanation: In an ideal throttling process, there is no work done ($W = 0$) and the process is assumed adiabatic with no heat transfer ($Q = 0$). By neglecting changes in kinetic and potential energies, the steady flow energy equation simplifies to $h_1 = h_2$. Thus, enthalpy remains constant.
Year: 2016 | Topic: Thermodynamics
Air at 100 kPa, 2°C occupies a 10 liter piston-cylinder device that is arranged to maintain constant air pressure. This device is now heated until its volume is 20 liters. The work produced by the air is
Answer: D. 1 kJ
Explanation: The process is isobaric (constant pressure). The work done by a gas during a constant pressure process is given by $W = P(V_2 - V_1)$. The pressure is $P = 100$ kPa = $100 \times 10^3$ Pa. The initial volume is $V_1 = 10$ liters = $10 \times 10^{-3}$ $m³$, and the final volume is $V_2 = 20$ liters = $20 \times 10^{-3}$ $m³$. Work done $W = 100 \times 10^3 \times (20 \times 10^{-3} - 10 \times 10^{-3}) = 100 \times 10^3 \times 10 \times 10^{-3} = 1000$ J = 1 kJ.
Year: 2016 | Topic: Thermodynamics
A heat engine working with a thermal efficiency of 35% receives 2 kW of heat from a furnace. The waste heat rejected from the engine is
Answer: C. 1.3 kW
Explanation: Thermal efficiency ($\eta$) is given by $1 - (Q_R / Q_S)$, where $Q_S$ is the heat supplied and $Q_R$ is the waste heat rejected. Given $\eta = 0.35$ and $Q_S = 2$ kW, we can write $0.35 = 1 - (Q_R / 2)$. Solving for $Q_R$ gives $Q_R / 2 = 1 - 0.35 = 0.65$, therefore $Q_R = 0.65 \times 2 = 1.3$ kW.
Year: 2016 | Topic: Thermodynamics
Joule-Thomson coefficient for an ideal gas is
Answer: C. zero
Explanation: The Joule-Thomson coefficient ($\mu$) represents the rate of change of temperature with respect to pressure during a throttling process (constant enthalpy). For a perfect ideal gas, enthalpy is a function of temperature only. Since enthalpy remains constant during throttling, the temperature also remains constant, resulting in a Joule-Thomson coefficient of strictly zero.
Year: 2016 | Topic: Thermodynamics
Air is heated from 0°C to 100°C in a sealed metal container. Its density
Answer: B. Decreases slightly
Explanation: Based on the provided solution text referencing ideal gas principles ($P = \rho RT$), it suggests that density decreases as temperature increases ($T \propto 1/\rho$). Note: In reality, for a perfectly rigid sealed container (constant volume and mass), density remains exactly the same, but the provided solution justifies it as "Decreases slightly" referencing proportional laws.
Year: 2016 | Topic: Thermodynamics
A system is undergoing a cycle that consists of three processes. During the first process, the work is 5 kJ and the heat is 23 kJ. In the second process no work takes place and the heat interaction is -50 kJ. The third process is adiabatic. The work in the third process is
Answer: A. -32 kJ
Explanation: According to the First Law of Thermodynamics for a complete cycle, the net heat transfer is equal to the net work transfer ($\Sigma Q = \Sigma W$). The total heat transfer is $Q_{net} = 23 - 50 + 0 = -27 \text{ kJ}$. The total work transfer is $W_{net} = 5 + 0 + W$. Equating the two gives $-27 = 5 + W$, solving for $W$ yields $W = -32 \text{ kJ}$.
Year: 2017 | Topic: Thermodynamics
Isochoric changes in ideal gas are characterized by
Answer: C. P/T = Const
Explanation: An isochoric process is a constant volume process. From the ideal gas law PV=nRT, if V is constant, then P/T = nR/V = constant.
Year: 2017 | Topic: Thermodynamics
The work ratio of a gas turbine is a function of
Answer: A. temperature ratio and pressure ratio
Explanation: The work ratio (Net Work / Turbine Work) of a Brayton cycle depends on both the pressure ratio and the temperature ratio (max temp / min temp).
Year: 2017 | Topic: Thermodynamics
Isothermal compressibility of an ideal gas in
Answer: B. nRT/VP²
Explanation: Isothermal compressibility is k_T = (-1/V)(∂V/∂P)_T. For an ideal gas, V = nRT/P. So, (∂V/∂P)_T = -nRT/P². Thus, k_T = (-1/V)(-nRT/P²) = nRT/(VP²).
Year: 2017 | Topic: Thermodynamics
Efficiency of gas turbines lies between
Answer: D. 15 to 18%
Explanation: Practical gas turbines, especially older or simple cycle designs, have a relatively low thermal efficiency, typically in the range of 15% to 25%.
Year: 2017 | Topic: Thermodynamics
The critical pressure ratio for maximum discharge in a steam nozzle is (where 'γ' is the isentropic index)
Answer: B. (2/(γ+1))^(γ/(γ-1))
Explanation: The critical pressure ratio (P*/P₀) for isentropic flow through a nozzle that maximizes the mass flow rate is given by the formula (2/(γ+1))^(γ/(γ-1)).
Year: 2017 | Topic: Thermodynamics
What is the maximum rate that a heat pump which uses 1 kW of electric power can supply heat to a house at 27°C when the outside temperature is 12°C
Answer: C. 20 kJ/s
Explanation: The maximum Coefficient of Performance (COP) for a heat pump is given by T_H / (T_H - T_L), where temperatures are in Kelvin. T_H = 27+273=300K, T_L = 12+273=285K. COP = 300 / (300-285) = 20. The rate of heat supply Q_H = COP * W_in = 20 * 1 kW = 20 kW = 20 kJ/s.
Year: 2017 | Topic: Thermodynamics
A jet engine consumes 1 kg of fuel for each 40 kg of air. Fuel consumption is 1 kg/sec. When aircraft travels at 200 m/sec, velocity of discharge gases is 700 m/sec. The power developed by the engine is
Answer: D. 4140 kW
Explanation: Thrust F = (ṁ_a + ṁ_f)V_e - ṁ_a*V_a = (40+1)*700 - 40*200 = 28700 - 8000 = 20700 N. Power = F * V_a = 20700 N * 200 m/s = 4,140,000 W = 4140 kW.
Year: 2017 | Topic: Thermodynamics
Which of the following is most appropriate for vapour power cycles?
Answer: A. In Carnot cycle, expansion of vapour occurs without change in entropy
Explanation: The expansion and compression processes in a Carnot cycle are isentropic (constant entropy). The other statements are incorrect descriptions of Rankine and Carnot cycles.
Year: 2017 | Topic: Thermodynamics
The reheat factor Rf for a steam turbine is usually in the range of
Answer: B. 1 to 1.065
Explanation: The reheat factor represents the increase in cumulative enthalpy drop due to reheating, which makes it slightly greater than 1, typically in the range of 1.02 to 1.06.
Year: 2017 | Topic: Thermodynamics
Isentropic process is related to
Answer: B. Reversible Adiabatic Expansion
Explanation: An isentropic process is one that is both adiabatic (no heat transfer) and reversible (no entropy generation). This combination results in a constant entropy process.
Year: 2017 | Topic: Thermodynamics
The temperature of the products of combustion when the maximum amount of chemical energy is converted to thermal energy is
Answer: C. Equal to adiabatic flame temperature
Explanation: The adiabatic flame temperature is the maximum possible temperature achieved for a given fuel and oxidizer, which occurs when the combustion is complete (stoichiometric) and there are no heat losses.
Year: 2018 | Topic: Thermodynamics
The pressure P of a gas in terms of its mean kinetic energy per unit volume E is equal to
Answer: D. 2E/3
Explanation: Kinetic energy for gas KE = (3/2)RT = (3/2)PV. Therefore, KE/V = (3/2)P = E, which means P = (2/3)E.
Year: 2018 | Topic: Thermodynamics
Work done in a free expansion process is
Answer: C. zero
Explanation: The work done in a free expansion process is zero because free expansion is an irreversible adiabatic and isentropic process. In free expansion, there is no external resistance upon which system can apply force.
Year: 2018 | Topic: Thermodynamics
If the value of n is infinitely large in a polytropic process PV^n=C, then the process is known as constant
Answer: A. volume
Explanation: PV^n = C can be written as P^(1/n) * V = C^(1/n). If n is infinity, then 1/n approaches 0, leaving V = Constant (C). This is a constant volume process.
Year: 2018 | Topic: Thermodynamics
During the charging of a storage battery, the current I is 20A and the voltage E in 12.8V. The ratio of heat transfer from the battery is 16W. At what rate is the internal energy increasing?
Answer: D. 240 Js^-1
Explanation: Energy stored in battery during charging = (V * I) - Q = (12.8 * 20) - 16 = 256 - 16 = 240 W (or Js^-1).
Year: 2018 | Topic: Thermodynamics
One kg of carbon monoxide requires how much quantity of oxygen to produce 11/7 kg of carbon dioxide gas?
Answer: A. 4/7 kg
Explanation: CO + 0.5 O2 -> CO2. 1/2 mole of O2 would make 1 mole CO2. So, 16 kg O2 would make 44 kg CO2. Then (16/44) kg O2 would make 1 kg CO2. Therefore, 11/7 kg CO2 is made by (11/7) * (16/44) kg O2 = 4/7 kg O2.
Year: 2018 | Topic: Thermodynamics
The entropy of water at 0 K is assumed to be
Answer: D. 0
Explanation: The entropy of water at 0 K is assumed to be zero.
Year: 2018 | Topic: Thermodynamics
An expression process as per law PV = constant is known as
Answer: B. hyperbolic expansion
Explanation: We know that PV = nRT = constant. For a constant number of moles, temperature is constant. In a P-V diagram, this represents a hyperbolic curve.
Year: 2018 | Topic: Thermodynamics
A simple compressible substance inside a cylinder undergoes a change of state quasistatically, isothermally and at constant pressure of 1.5 MPa. If the enthalpy change for the system is 4000 kJ at 150 °C, what is the corresponding change in its entropy?
Answer: A. 9.45 kJ K^-1
Explanation: Since it is a phase change (temperature is constant), we use dH = TdS + VdP. At constant pressure (dP = 0), dH = TdS. Entropy change dS = dH / T = 4000 / (150 + 273) = 9.45 kJ/K.
Year: 2018 | Topic: Thermodynamics
For a two stage reciprocating compressor, compression from P1 to P3 is with perfect inter cooling and no pressure losses. If compression in both the cylinders follows the same polytropic process and the atmospheric pressure is Pa, then the intermediate pressure P2 is given by
Answer: B. P2 = $\sqrt{P1 * P3}$
Explanation: For perfect inter-cooling, intermediate pressure P2 is the geometric mean of initial and final pressures: P2 = $\sqrt{P1 * P3}$.
Year: 2024 | Topic: Thermodynamics
Vapor pressure in mm of mercury of solid and liquid ammonia is given by $\ln P = 23.1 - (3749/T)$ and $\ln P = 19.6 - (3049/T)$, find out the triple point.
Answer: C. 200 K
Explanation: At the triple point, the vapor pressures of the solid and liquid phases are equal. Equating the two expressions: $23 Jan 3749/T = 19 Jun 3049/T$. Rearranging gives $23.1 - 19.6 = 3749/T - 3049/T$, so $3.5 = 700/T$. Solving for $T$ yields $T = 700 / 3.5 = 200$ K.
Year: 2024 | Topic: Thermodynamics
What is the change in entropy when water is vaporized at 100 °C. Latent heat of vaporization of water may be taken as 2238 kJ/kg.
Answer: C. 6 kJ/kg K
Explanation: The change in entropy dS during phase change at constant temperature is dQ / T. Temperature T = 100 + 273 = 373 K. Latent heat dQ = 2238 kJ/kg. Therefore, dS = 2238 / 373 = 6 kJ/kg K.
Year: 2024 | Topic: Thermodynamics
Examine the following statements: (i) Work is high grade energy. (ii) It is possible for a heat engine to produce work in a cycle by exchanging heat with bodies at same temperature. (iii) It is impossible to construct a device which operating in a cycle will produce no other effect other than transfer of heat from a cooler to a hotter body.
Answer: C. Statements (i) and (iii) are correct
Explanation: Statement (i) is true; work is fully convertible to heat, making it high-grade energy. Statement (ii) is false according to the Kelvin-Planck statement, which requires interaction with at least two thermal reservoirs at different temperatures. Statement (iii) is the Clausius statement of the Second Law of Thermodynamics, which is true.
Year: 2024 | Topic: Thermodynamics
Molliar diagram is also called as
Answer: D. Enthalpy entropy chart
Explanation: A Mollier diagram is a thermodynamic diagram representing the enthalpy (h) of a system versus its entropy (s), hence it is called an Enthalpy-entropy chart.
Year: 2024 | Topic: Thermodynamics
Real gas behavior approaches ideal gas equation when
Answer: D. Pressure is very low and temperature is very high
Explanation: Real gases behave most like ideal gases at very low pressures and high temperatures because the intermolecular forces become negligible and the volume occupied by the gas molecules is extremely small compared to the total volume.
Year: 2024 | Topic: Thermodynamics
For water, the latent heat of freezing is 334 kJ/kg and the specific heat capacity averages 4.19 kJ/kgK. The quantity of heat to be removed from 1 kg of water at 30 °C in order to turn it into ice at 0 °C is:
Answer: D. 459 kJ
Explanation: Total heat removed Q = Sensible heat + Latent heat. Sensible heat to cool water from 30 °C to 0 °C = m * C * dT = 1 * 4.19 * 30 = 125.7 kJ. Latent heat to freeze = m * L = 1 * 334 = 334 kJ. Total Q = 125.7 + 334 = 459.7 kJ, which is approximately 459 kJ.
Year: 2024 | Topic: Thermodynamics
Which of the following represent sterling cycle
Answer: B. I and III
Explanation: A Stirling cycle consists of two isothermal processes and two isochoric (constant volume) processes. Graph I shows this on a P-V diagram (two curves for isothermal and two vertical lines for constant volume). Graph III shows this on a T-S diagram (two horizontal lines for isothermal and two curves for constant volume).