Question

When one mole of mono-atomic ideal gas at T K undergoes adiabatic change under a constant external pressure of 1 atm change volume from 1 litre to 2 litre. The final temperature in Kelvin would be:

• A. $\frac{T}{2^{\left(2/3\right)}}$
• B. $T+\frac{2}{3\times 0.0821}$
• C. $T$
• D. $T-\frac{2}{3\times 0.0821}$

Answer 1

Answer: (D)

$T-\frac{2}{3\times 0.0821}$

Solution:- (D) $T-\frac{2}{3\times 0.0821}$

Since the process is adiabatic,

$dq=0$

From ideal gas equation,

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

As we know that,

$W=-P_{ext.}\Delta V$

$\Delta U=n{C}_{v}dT$

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

$n{C}_{v}dT=-P_{ext.}\Delta V.....\left(2\right)$

Given:-

$n=1\text{mole}$

$T_i=T$

$T_f=?$

$V_i=1L$

$V_f=2L$

$P_{ext.}=1atm$

$C_v=\frac{3}{2}R$

$R=0.0821L-atm{mol}^{-1}{K}^{-1}$

Now from ${eq}^n\left(2\right)$, we have

$1\times \frac{3}{2}R\times (T_f-T)=-1(2-1)$

$\Rightarrow T_f-T=-\frac{2}{3R}$

$\Rightarrow T_f=T-\frac{2}{3\times 0.0821}$

Hence the final temperature will be $(T-\frac{2}{3\times 0.0821})$.