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Heat and Thermodynamics question

2019 · 10 Apr · Shift 1 · Q67
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Heat and Thermodynamics question

2019 · 10 Apr · Shift 1 · Q67

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
n moles of an ideal gas with constant volume heat capcity CV undergo an isobaric expansion by certain volume. The ratio of the work done in the process, to the heat supplied is :
  1. A
    nRCV−nR{{nR} \over {{C_V} - nR}}CV​−nRnR​
  2. B
    4nRCV−nR{{4nR} \over {{C_V} - nR}}CV​−nR4nR​
  3. C
    4nRCV+nR{{4nR} \over {{C_V} + nR}}CV​+nR4nR​
  4. D
    nRCV+nR{{nR} \over {{C_V} + nR}}CV​+nRnR​
View written solutionFree

Correct answer: D

  1. Given: An ideal gas undergoes an isobaric expansion.

  2. For an isobaric process, the work done is W=PΔVW = P\Delta VW=PΔV

    Using the ideal gas law, PΔV=nRΔTP\Delta V = nR\Delta TPΔV=nRΔT So, W=nRΔTW = nR\Delta TW=nRΔT

  3. Heat supplied in an isobaric process is Q=nCPΔTQ = nC_P\Delta TQ=nCP​ΔT

  4. Since the question gives constant volume heat capacity CVC_VCV​, use CP=CV+nRC_P = C_V + nRCP​=CV​+nR

    Here CVC_VCV​ is the heat capacity of the given nnn moles, so this relation is appropriate.

  5. Therefore, Q=(CV+nR)ΔTQ = (C_V + nR)\Delta TQ=(CV​+nR)ΔT

  6. Now the required ratio is WQ=nRΔT(CV+nR)ΔT\frac{W}{Q} = \frac{nR\Delta T}{(C_V + nR)\Delta T}QW​=(CV​+nR)ΔTnRΔT​

    Cancelling ΔT\Delta TΔT, WQ=nRCV+nR\frac{W}{Q} = \frac{nR}{C_V + nR}QW​=CV​+nRnR​

  7. This matches Option D.

Therefore, the correct answer is: nRCV+nR\boxed{\frac{nR}{C_V+nR}}CV​+nRnR​​

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