Question

The volume of a gas is reduced to $\frac{1}{4}$ of its initial volume adiabatically at ${27}^o$. The final temperature of the gas (if $\gamma =1.4$) will be

• A. $27\times (4{)}^{0.4}K$
• B. $300\times (1/4{)}^{0.4}K$
• C. $100\times (4{)}^{0.4}K$
• D. $300\times (4{)}^{0.4}K$

Answer 1

The correct option is D $300\times (4{)}^{0.4}K$

The equation of state for an adiabatic process:
$T{V}^{\gamma -1}=constant$
i.e. $T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$

Given:
$T_1=300K$, $V_2=\frac{V_1}{4}$

$T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$
$300\times V_1^{0.4}=T_2\frac{V_1^{0.4}}{4}$
$T_2=300\times (4{)}^{0.4}$