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Heat and Thermodynamics question

2007 · Shift 0 · Q93
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Heat and Thermodynamics question

2007 · Shift 0 · Q93

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
If Cp{C_p}Cp​ and Cv{C_v}Cv​ denote the specific heats of nitrogen per unit mass at constant pressure and constant volume respectively, then
  1. A
    Cp−Cv=28R{C_p} - {C_v} = 28RCp​−Cv​=28R
  2. B
    Cp−Cv=R/28{C_p} - {C_v} = R/28Cp​−Cv​=R/28
  3. C
    Cp−Cv=R/14{C_p} - {C_v} = R/14Cp​−Cv​=R/14
  4. D
    Cp−Cv=R{C_p} - {C_v} = RCp​−Cv​=R
View written solutionFree

Correct answer: B

  1. For an ideal gas, the relation between molar specific heats is Cp,m−Cv,m=RC_{p,m} - C_{v,m} = RCp,m​−Cv,m​=R where RRR is the universal gas constant.

  2. But in the question, CpC_pCp​ and CvC_vCv​ are specific heats per unit mass (not per mole).

  3. For specific heats per unit mass, the corresponding relation is Cp−Cv=RMC_p - C_v = \frac{R}{M}Cp​−Cv​=MR​ where MMM is the molar mass of the gas in kg/mol.

  4. For nitrogen gas N2N_2N2​, M=28 g/mol=28×10−3 kg/molM = 28\,\text{g/mol} = 28 \times 10^{-3}\,\text{kg/mol}M=28g/mol=28×10−3kg/mol

  5. Therefore, Cp−Cv=R28×10−3C_p - C_v = \frac{R}{28 \times 10^{-3}}Cp​−Cv​=28×10−3R​ If one uses gram-based mass units as implied by the options, this is written as Cp−Cv=R28C_p - C_v = \frac{R}{28}Cp​−Cv​=28R​

  6. Hence the correct option is B\boxed{\text{B}}B​

Option check

  • A: 28R28R28R — incorrect
  • B: R/28R/28R/28 — correct
  • C: R/14R/14R/14 — incorrect
  • D: RRR — incorrect
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