Question

When one mole of monoatomic ideal gas at T K undergoes adiabatic change under a constant external pressure of $1$ atm changes volume from $1$ litre to $2$ litres. 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. $\frac{T}{2^{2/3}}$.

For an adiabatic process,

$T{V}^{\gamma -1}=\text{constant}$

Therefore,

$T_1{V_1}^{\gamma -1}=T_2{V_2}^{\gamma -1}$

$\because \gamma =\frac{5}{3}(\because \text{the gas is monoatomic})$

$\therefore T_1{V_1}^{\frac{5}{3}-1}=T_2{V_2}^{\frac{5}{3}-1}$

$\Rightarrow T_1{V_1}^{\frac{2}{3}}=T_2{V_2}^{\frac{2}{3}}$

Given that:-

$V_1=1L$

$V_2=2L$

$T_1=TK$

$\therefore T{\left(1\right)}^{\frac{2}{3}}=T_2{\left(2\right)}^{\frac{2}{3}}$

$\Rightarrow T_2=\frac{T}{2^{\frac{2}{3}}}$

Hence the final temperature will be $\frac{T}{2^{2/3}}$.