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Question

Two moles of an ideal monoatomic gas occupies a volume V at $$27^o$$C. The gas expands adiabatically to a volume $$2$$V. Calculate (a) the final temperature of the gas and (b) change on its internal energy.


A
(a) 189K (b) 2.7 kJ
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B
(a) 195K (b) 2.7 kJ
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C
(a) 189K (b) 2.7kJ
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D
(a) 195k (b) 2.7kJ
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Solution

The correct option is A (a) $$189$$K (b) $$-2.7$$ kJ
In an adiabatic process, $$PV^{ \gamma} = constant$$

or $$T V^{\gamma -1 } = constant$$

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

For monoatomic gas
$$ \gamma = \dfrac{5}{3}$$
$$T_2 = (300K) (\dfrac{V}{2V})^{\dfrac{5}{3} - 1}$$
$$= 189 K$$

Change in internal energy 

$$\Delta U = n C_v \Delta T$$

 $$= 2 (\dfrac{3}{2})(\dfrac{25}{3})(-111) $$

$$= -2.7 kJ$$

Physics

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