A diatomic ideal gas initially at 273K is given 100cal heat due to which system did 209J work. Molar heat capacity (Cm) of gas for the process is:
A
32R
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B
52R
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C
54R
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D
5R
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Solution
The correct option is D5R From first law of thermodynamics : ΔU = q+W
and △U=nCv△T
so, nCv△T=100×4.2−209=420−209=211Jn×5R2×△T=211n△T=211×25R
molar heat capacity is given as: Cm=qn△T=420211×2×5R≈5R