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Heat and Thermodynamics question

2021 · 26 Feb · Shift 2 · Q67
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Heat and Thermodynamics question

2021 · 26 Feb · Shift 2 · Q67

JEE MainPhysicsHeat and ThermodynamicsNumerical+4 / −1
1 mole of rigid diatomic gas performs a work of Q5{Q \over 5}5Q​ when heat Q is supplied to it. The molar heat capacity of the gas during this transformation is xR8{xR \over 8}8xR​. The value of x is ‾\underline{\hspace{2cm}}​. [R = universal gas constant]
Numerical answer
View written solutionFree

Correct answer: 25

  1. Given data
  • Number of moles: n=1n=1n=1
  • Gas is rigid diatomic? Here “rigid diatomic gas” means a diatomic ideal gas with no vibrational modes active, so CV=5R2C_V=\frac{5R}{2}CV​=25R​
  • Heat supplied: QQQ
  • Work done by gas: W=Q5W=\frac{Q}{5}W=5Q​

We need the molar heat capacity during this process.


  1. Use the first law of thermodynamics

For the heat supplied, Q=ΔU+WQ=\Delta U+WQ=ΔU+W So, ΔU=Q−W=Q−Q5=4Q5\Delta U=Q-W=Q-\frac{Q}{5}=\frac{4Q}{5}ΔU=Q−W=Q−5Q​=54Q​


  1. Relate internal energy change to temperature change

For 1 mole of a diatomic ideal gas, ΔU=nCVΔT=5R2ΔT\Delta U=nC_V\Delta T=\frac{5R}{2}\Delta TΔU=nCV​ΔT=25R​ΔT Thus, 5R2ΔT=4Q5\frac{5R}{2}\Delta T=\frac{4Q}{5}25R​ΔT=54Q​

Solve for ΔT\Delta TΔT: ΔT=4Q5⋅25R=8Q25R\Delta T=\frac{4Q}{5}\cdot \frac{2}{5R}=\frac{8Q}{25R}ΔT=54Q​⋅5R2​=25R8Q​


  1. Find the molar heat capacity for the process

By definition, C=QnΔTC=\frac{Q}{n\Delta T}C=nΔTQ​ Since n=1n=1n=1, C=QΔT=Q8Q25R=25R8C=\frac{Q}{\Delta T}=\frac{Q}{\frac{8Q}{25R}}=\frac{25R}{8}C=ΔTQ​=25R8Q​Q​=825R​

Given that C=xR8C=\frac{xR}{8}C=8xR​ we compare: xR8=25R8\frac{xR}{8}=\frac{25R}{8}8xR​=825R​ Hence, x=25x=25x=25


  1. Comparison with stored answer

Derived answer: 252525

Stored correct answer: 252525

They match.

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