A diatomic ideal gas initially at 273k is given 100 cal heat due to which system did 209 J work Molar heat capacity $(C_{m})$ of gas for the process is :

\frac{3}{2}

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\frac{5}{2}

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\frac{5}{4}

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5 R

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Solution

The correct option is D 5 R
Given : Diatomic molecule at 273K
'q' absorbed = positive = +100Cal = 100 x 4.184J = 418.4J
'W' done by system = negative = -209J
By first law of thermodynamics;
$Δ$ U = q + W = 418.4 + (-209) = 209.4J
We know for diatomic molecule $C_{v} = \frac{5}{2} R$ and $C_{v} Δ T = Δ U$
$C_{v} Δ T = 209.4$
$\frac{5}{2} R Δ T = 209.4$
$Δ T = \frac{209.4 \times 2}{5 R}$
And, $H e a t e x c h a n g e = C_{m} \times Δ T$
where; $C_{m}$ is molar heat capacity
$C_{m} = \frac{H e a t E x c h a n g e}{Δ T}$
substituting values for Heat Exchange = 418.4 and $Δ T = \frac{209.4 \times 2}{5 R}$
$C_{m} = 5 R$

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A diatomic ideal gas initially at 273k is given 100 cal heat due to which system did 209 J work Molar heat capacity (Cm) of gas for the process is :

A diatomic ideal gas initially at 273k is given 100 cal heat due to which system did 209 J work Molar heat capacity $(C_{m})$ of gas for the process is :