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Question

Calculate the work done by 10 g of an ideal gas of molecular weight 44 g in expanding reversibly and isothermaly from a volume of 5 L to 10 L at 300 K.

A
w = 103 cal
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B
w = 94 cal
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C
w = 9.4 cal
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D
None of these
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Solution

The correct option is B w = 94 cal
Work done in an isothermal reversible process is given as:
w=2.303nRT log10V2V1
Number of moles of ideal gas n = 10 g44 g mol1 = 1044 mol

R = 2 cal K1 mol1
T = 300 K
V2V1= 105 = 2

On substituting the values we get,
w = 2.303×1044 × 2 × 300 × log 2
On solving,
w = 93.93 cal

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