Question

When one mole of monoatomic ideal gas at $T\text{K}$ undergoes adiabatic change under a constant external pressure of $1\text{atm}$, volume changes from $1\text{litre}$ to $2\text{litre}$. The final temperature in Kelvin would be:

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

Answer 1

The correct option is A $\frac{T}{2^{2/3}}$

Since here we are taking an adibatic process, thus, $T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$
For a monoatomic gas, $\gamma =\frac{5}{3}$
$\therefore T_1{V}_1^{\frac{5}{3}-1}=T_2{V}_2^{\frac{5}{3}-1}$

Given, $\text{Initial volume}{V}_1=1\text{litre}$
$\text{Initial temperature}=T$
$\text{Final volume}{V}_2=2\text{litre}$
$\therefore T(1{)}^{\frac{2}{3}}=T_2(2{)}^{\frac{2}{3}}$
$\Rightarrow T_2=\frac{T}{2^{2/3}}$