Question

When one mole of monoatomic ideal gas at temperature $T$ undergoes adiabatic change reversibly, change in volume is from $1Lto2L$, 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 $\frac{T}{2^{2/3}}$

We know the relation for an adiabatic reversible process :
$\frac{T_1}{T_2}={\left(\frac{V_2}{V_1}\right)}^{\gamma -1}$
where, $T_1=$ initial temperature
$T_2=$ final temperature
$V_1=$ initial volume
$V_2=$ final volume
$\gamma =\frac{C_p}{C_v}=\frac{5}{3}$ (for monoatomic gas)
so,
$\frac{T_1}{T_2}={\left(\frac{2}{1}\right)}^{\left(5/3\right)-1}=2^{\left(2/3\right)}$
$T_{{}_2}=\frac{T_1}{2^{\left(2/3\right)}}$