Question

An ideal monoatomic gas $(C_v=1.5R)$ initially at $298$ K and $1.013$ atm expands adiabatically irreversibly until it is in equilibrium with a constant external pressure of $0.1013$ atm. The final temperature (in Kelvin) of the gas is :

• A. $190.7$
• B. $119.7$
• C. $278$
• D. $273$

Answer 1

The correct option is A $190.7$

We know the relation, $C_v=\frac{R}{\gamma -1}$, where $\gamma =\frac{C_p}{C_v}$ and $C_p-C_v=R$

$P{V}^{\gamma }=$ constant for adiabatic process (reversible)

$W=C_v(T_2-T_1)=-R{P}_{ext}(\frac{T_2}{P_2}-\frac{T_1}{P_1})$

Substituting the values, we get-

$1.5\times R\times (-298+T_2)=-R\times \left(0.1013\right)(\frac{T_2}{0.1013}-\frac{298}{1.013})$

$⟹$ $1.5\times R\times (T_2-298)=-R\times (T_2-29.8)$

$⟹$$1.5\times T_2-1.5\times 298=-T_2+29.8$

$⟹$ $2.5\times T_2=1.6\times 298$

$⟹$ $T_2=\frac{1.6\times 298}{2.5}$ $=190.72K$

Option A is correct.