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.
- A. p, q
- B. r, s
- C. p, r, t
- D. q, t
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.