Question

One mole of monoatomic ideal gas $T\left(K\right)$ undergoes reversible adiabatic change under a constant external pressure of 1 atm. Changes volume from 1 litre to 2 litre. The final temperature is:

• A. $T+\frac{2}{3\times R}$
• B. $T$
• C. $T-\frac{3}{2R}$
• D. $\frac{T}{2^{2/3}}$

Answer 1

The correct option is C $T-\frac{3}{2R}$

For adiabatic condition, $dq=0$

Thus by first law of thermodynamics.

$dU=dW+dq$

$\Rightarrow dU=dW⟶\left(1\right)$ as $dq=0$ for adiabatic

$\Rightarrow dW=-P_{ext}(V_2-V_1)$

$\Rightarrow dU=CvdT$

For monoatomic gas

Degree of freedom= $3M=3\times 1=3⟶$ All transitional

$C_v=\frac{3}{2}K$ or for $1$ mole $K=R$

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

Putting all values in equation $1$, we have

$C_{v}dT=-P_{ext}(V_2-V_1)$

$\Rightarrow \frac{3}{2}(T_2-T_1)=-1(2L-1L)$

$\Rightarrow (T_2-T_1)=-1\times \frac{2}{3R}$

$\Rightarrow T_2=T_1-\frac{2}{3R}$

Thus, final temperature is $T-\frac{2}{3R}$ .