Question

A diatomic gas initially at $18^{\circ }C$ is compressed adiabatically to one eighth of its original volume. The temperature after compression will be

• A. $395.4^{\circ }C$
• B. $144^{\circ }C$
• C. $18^{\circ }C$
• D. $887.4^{\circ }C$

Answer 1

Answer: (A)

Initial temperature $(T_1)=18^{0}C$
= (273 + 18) = 291 K and $V_2=(1/8)V_1.$
For adiabatic compression, $T{V}^{\gamma -1}=constant$
or $T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}.$
Therefore $T_2=T_1(\frac{V_1}{V_2}{)}^{\gamma -1}$
$=291\times (8{)}^{1.4-1}=291\times (8{)}^{0.4}$
$=291\times 2.297=668.4K=395.4^{0}C.$