Question

One mole of ideal gas is allowed to expand reversibly and adiabatically from a temperature of ${27}^o$C. If the work done by the gas in the process is $3$ kJ, the final temperature will be equal to ($C_v=20$ J/K mol):

• A. $100$ K
• B. $450$ K
• C. $150$ K
• D. $400$ K

Answer 1

The correct option is C $150$ K

Solution:- (C) $150K$

From first law of thermodynamics,

$\Delta U=q+W$

As we know that, for adiabatic condition,

$q=0$

$\therefore \Delta U=W.....\left(1\right)$

At constant volume,

$\Delta U=n{C}_v\Delta T.....\left(2\right)$

From ${eq}^n\left(1\right)\&\left(2\right)$, we have

$W=n{C}_v\Delta T.....\left(3\right)$

Given:-

$T_i=27℃=(27+273)K=300K$

$T_f=T\left(\text{say}\right)=?$

$\Delta T=T_f-T_i=(T-300)$

$C_v=20J/K-mol$

$W=-3kJ=-3\times {10}^{3}J[\because \text{Work done by the gas is negative}]$

Substituting all these values in ${eq}^n\left(3\right)$, we have

$-3000=1\times 20\times (T-300)$

$T-300=-150$

$\Rightarrow T=300-150=150K$

Hence the final temperature is $150K$.