Question

When one mole of monatomic ideal gas at t k undergoes adiabatic change under a constant external pressure of 1 atm changes volume from 1 litre to 2 litre. The final temperature in kelvin would be?

Answer 1

use formula,
$C_V(T_2-T_1) = P_{ext}(V_1-V_2)$

we know, $C_P-C_V=R$
$\implies\frac{C_P}{C_V}- 1=\frac{R}{C_V}$
$\Rightarrow \gamma - 1=\frac{R}{C_V}$

$C_V = \frac{R}{\gamma -1}$

for monochromatic,
$\gamma=\frac{5}{3}$

now,
$\frac{R}{\gamma-1}(T_f-T)=1atm(1L-2L)\\\\\frac{R}{\gamma-1}(T_f-T)=-1\\\\T_f-T=\frac{1-\gamma}{R}\\\\T_f=T-\frac{\frac{2}{3}}{0.0821}$

hence, final temperature is
(T - 2/3 × 0.0821 )K