Question

One mol of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of ${27}^{o}C$. If the work done during the process is 3 kJ, then the final temperature of gas is:
Given: $C_v$ for the gas = $20J/K$

• A. 100 K
• B. 150 K
• C. 195 K
• D. 255 K

Answer 1

The correct option is B 150 K

The gas expands adiabatically, so the heat transfer is zero.
From the first law, $\Delta U=w$
At constant volume, $\Delta U=n{C}_{v}dT$
So, w = $n{C}_{v}dT$
Given, $T_1=300K$ ; $T_2$ = T K
Since, the work done is by the gas w = $-3\times {10}^3$ kJ
For the gas $C_v=20J/K$
$\therefore$ $-3000$ = $1\times 20\times (T-300$)
which gives T = 150 K