A diatomic ideal gas is heated at constant volume until the pressure is doubled and again heated at constant pressure until the volume is doubled. The average molar heat capacity for the whole process is

\frac{13 R}{6}

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\frac{19 R}{6}

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\frac{23 R}{6}

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\frac{17 R}{6}

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Solution

The correct option is B $\frac{19 R}{6}$
For the isochoric process we have
$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$
Thus we get $T_{2} = 2 T_{1}$

For isobaric process
$\frac{V_{2}}{T_{2}} = \frac{V_{3}}{T_{3}}$
Thus we get
$T_{3} = 2 T_{2} = 4 T_{1}$
Thus for the entire process we have $Δ T = 4 T_{1} - T_{1} = 3 T_{1}$

For the isochoric process we have heat absorbed as $n C_{v} Δ T = n \frac{5}{2} R (2 T_{1} - T_{1})$ and for isobaric process we have $n C_{p} Δ T = n (1 + \frac{5}{2}) R (4 T_{1} - 2 T_{1})$
Thus we get
$n C Δ T = n (\frac{5}{2} R) T_{1} + n (\frac{7}{2} R) 2 T_{1}$
Solving we get C as $\frac{19}{6} R$

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