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Heat and Thermodynamics question

2021 · 25 Feb · Shift 1 · Q58
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  5. /2021 · 25 Feb · Shift 1 · Q58

Heat and Thermodynamics question

2021 · 25 Feb · Shift 1 · Q58

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
A diatomic gas, having Cp=72R{C_p} = {7 \over 2}RCp​=27​R and Cv=52R{C_v} = {5 \over 2}RCv​=25​R, is heated at constant pressure. The ratio dU : dQ : dW :
  1. A
    5 : 7 : 3
  2. B
    3 : 7 : 2
  3. C
    5 : 7 : 2
  4. D
    3 : 5 : 2
View written solutionFree

Correct answer: C

  1. Use the first law of thermodynamics

For a small amount of heat supplied at constant pressure, dQ=dU+dWdQ = dU + dWdQ=dU+dW

Also,

  • dQ=nCp dTdQ = nC_p\,dTdQ=nCp​dT
  • dU=nCv dTdU = nC_v\,dTdU=nCv​dT
  • dW=P dVdW = P\,dVdW=PdV

For an ideal gas at constant pressure, dW=nR dTdW = nR\,dTdW=nRdT

  1. Substitute the given heat capacities

Given: Cp=72R,Cv=52RC_p = \frac{7}{2}R, \qquad C_v = \frac{5}{2}RCp​=27​R,Cv​=25​R

So, dU=n(52R)dTdU = n\left(\frac{5}{2}R\right)dTdU=n(25​R)dT dQ=n(72R)dTdQ = n\left(\frac{7}{2}R\right)dTdQ=n(27​R)dT dW=nR dTdW = nR\,dTdW=nRdT

  1. Write the ratio

Therefore, dU:dQ:dW=52:72:1dU : dQ : dW = \frac{5}{2} : \frac{7}{2} : 1dU:dQ:dW=25​:27​:1

Multiply by 2: dU:dQ:dW=5:7:2dU : dQ : dW = 5 : 7 : 2dU:dQ:dW=5:7:2

  1. Check options
  • A: 5:7:35:7:35:7:3 ❌
  • B: 3:7:23:7:23:7:2 ❌
  • C: 5:7:25:7:25:7:2 ✅
  • D: 3:5:23:5:23:5:2 ❌

Hence, the correct answer is Option C.

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