Question

The volume of a gas is reduced adiabatically to $\frac{1}{4}$ of its volume at ${27}^{o}C$. If $\gamma =1.4$. The new temperature will be:

• A. $300\times {\left(4\right)}^{0.4}K$
• B. $150\times {\left(4\right)}^{0.4}K$
• C. $250\times {\left(4\right)}^{0.4}K$
• D. none of these

Answer 1

Answer: (A)

$300\times {\left(4\right)}^{0.4}K$

For adiabatic change, the equation is
$T{V}^{\gamma -1}=$ constant
$T_1{V}_1^{\gamma -1}-T_2{V}_2^{\gamma -1}$
$(27+273)V_1^{\gamma -1}=T_2{\left(\frac{V_1}{4}\right)}^{\gamma -1}$
$300\times V_1^{\gamma -1}=\frac{T_2\times V_1^{\gamma -1}}{4^{\gamma -1}}\Rightarrow T_2=300\times 4^{\gamma -1}$
$T_2=300\times 4^{1.4-1}=300\times 4^{0.4}K$

Hence, option A is correct