Question

One mole of gas of specific heat ratio $1.5$ being initially at temperature $290K$ is adiabatically compressed to increase its pressure $8$ times. The temperature of the gas after compression will be

• A. $580K$
• B. $870K$
• C. $290\sqrt{2}K$
• D. $1160K$

Answer 1

Answer: (A)

$580K$

Given,

$T_1=290K$

$\gamma =1.5$

$P_2=8P_1$

In adiabatic process,

$P{V}^{\gamma }=k\left(constant\right)$

$P_1{V}_1^{\gamma }=P_2{V}_1^{\gamma }$ . . . . . .(1)

From ideal gas equation,

$PV=nRT$

$V=\frac{nRT}{P}$ . . . . . . .(2)

Substitute equation (2)

$P_1(\frac{nR{T}_1}{P_1}{)}^{\gamma }=P_2(\frac{nR{T}_2}{P_2}{)}^{\gamma }$

$P_1^{1-\gamma }(nR{T}_1{)}^{\gamma }=P_2^{1-\gamma }(nR{T}_2{)}^{\gamma }$

$P_1^{1-\gamma }{T}_1^{\gamma }=P_2^{1-\gamma }{T}_2^{\gamma }$

$P_1^{1-1.5}(290{)}^{1.5}=(8P_1{)}^{1-1.5}{T}_2^{1.5}$

$(\frac{P_1}{8P_1}{)}^{-1/2}=(\frac{T_2}{290}{)}^{1.5}$

$\frac{T_2}{290}=2$

$T_2=2\times 290=580K$

The correct option is A.