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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
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B
769 J
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C
1373 J
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D
969 J
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Solution

The correct option is A 1373 J
Given,
$$\gamma=1.4$$
$$n=1$$
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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