Question

1 mole of an ideal gas STP is subjected to a reversible adiabatic expansion to double its volume. The change in internal energy $(\gamma =1.4)$

• A. 1169 J
• B. 769 J
• C. 1373 J
• D. 969 J

Answer 1

Answer: (C)

1373 J

Given,

$\gamma =1.4$

$n=1$

At STP, $T_1=273.15K$

$V_1=V$

$V_2=2V$

In adiabatic process,

$P{V}^{\gamma }=constant=k$. . . . . . . .(1)

From ideal gas equation,

$PV=nRT$

$P=\frac{1\times RT}{V}=\frac{RT}{V}$. . . . .. (2)

Substitute equation (2) in equation (1), we get

$\frac{RT}{V}{V}^{\gamma }=k$

$T{V}^{\gamma -1}=k^{\prime }\left(constant\right)$

$T_1{V}_1^{\gamma -1}=T_2{V}_2^{\gamma -1}$

$T_2=T_1(\frac{V_1}{V_2}{)}^{\gamma -1}$

$T_2=273.15\times (\frac{V}{2V}{)}^{1.4-1}$

$T_2=207.00K$

Change in internal energy for adibatic process,

$\Delta U=\frac{R}{\gamma -1}(T_1-T_2)$

$\Delta U=\frac{8.31}{1.4-1}(273.15-207.00)$

$\Delta U=1373J$

The correct option is C.