Question

An amount of 1 mole of a gas is changed from its initial state (20 L, 2 atm) to final state (4 L, 10 atm), respectively. If the change can be represented by a straight line in $P-V$ curve, then the maximum temperature achieved by the gas in the process is ($R=0.08$ L-atm/K-mol)

• A. $900^\circ$C
• B. 627 K
• C. 900 K
• D. $1173^\circ$C

Answer 1

Answer: (C)

900 K

Answer 2

The process follows a linear path $P = -\frac{1}{2}V + 12$ in the P-V diagram. Temperature $T = \frac{PV}{nR}$. Substituting $P$, we get $T \propto -\frac{1}{2}V^2 + 12V$. This quadratic in $V$ is maximized at $V = -\frac{12}{2(-1/2)} = 12$ L, which is within the process range. At $V=12$ L, $P=6$ atm, giving $(PV)_{\text{max}} = 72$ L-atm. Thus, $T_{\text{max}} = \frac{72}{1 \times 0.08} = 900$ K.