Question

When one mole of a monoatomic ideal gas at T K undergoes adiabatic change under constant external pressure of 1atm changes from 1L to 2L volume. The final temperature of the gas in Kelvin is

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

Answer 1

Answer: (B)

. $\frac{T}{(2{)}^{\frac{2}{3}}}$.

For Adiabatic process:

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

$\therefore T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$

For a monoatomic gas, $\gamma =\frac{5}{3}$
$\Rightarrow T_1{V}_1^{\frac{5}{3}-1}=T_2{V}_2^{\frac{5}{3}-1}$
$\Rightarrow T(1{)}^{\frac{5}{3}-1}=T_2(2{)}^{\frac{5}{3}-1}$
$\Rightarrow T_2=\frac{T}{2^{\frac{5}{3}-1}}=\frac{T}{2^{\frac{2}{3}}}$.