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

Solution

## The correct option is A 1373 JGiven,$$\gamma=1.4$$$$n=1$$At STP, $$T_1=273.15K$$$$V_1=V$$$$V_2=2V$$In adiabatic process,  $$PV^{\gamma}=constant=k$$. . . . . . . .(1)From ideal gas equation,$$PV=nRT$$$$P=\dfrac{1\times RT}{V}=\dfrac{RT}{V}$$. . . .  .. (2)Substitute equation (2) in equation (1), we get$$\dfrac{RT}{V}V^{\gamma}=k$$$$TV^{\gamma-1}=k'(constant)$$$$T_1V_1^{\gamma-1}=T_2V_2^{\gamma-1}$$$$T_2=T_1(\dfrac{V_1}{V_2})^{\gamma-1}$$$$T_2=273.15\times (\dfrac{V}{2V})^{1.4-1}$$$$T_2=207.00K$$Change in internal energy for adibatic process,$$\Delta U=\dfrac{R}{\gamma-1}(T_1-T_2)$$$$\Delta U=\dfrac{8.31}{1.4-1}(273.15-207.00)$$$$\Delta U=1373J$$The correct option is C.Chemistry

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