JEE Challenger
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Comparison of Molar Specific Heat Capacities for Diatomic and Monoatomic Gases

Cv\text{C}_\text{v} and Cp\text{C}_\text{p} denote the molar specific heat capacities of a gas at constant volume and constant pressure, respectively. Then

Options

A

Cp−Cv\text{C}_\text{p} - \text{C}_\text{v} is larger for a diatomic ideal gas than for a monoatomic ideal gas

B

Cp+Cv\text{C}_\text{p} + \text{C}_\text{v} is larger for a diatomic ideal gas than for a monoatomic ideal gas

Correct
C

Cp/Cv\text{C}_\text{p} / \text{C}_\text{v} is larger for a diatomic ideal gas than for a monoatomic ideal gas

D

Cp⋅Cv\text{C}_\text{p} \cdot \text{C}_\text{v} is larger for a diatomic ideal gas than for a monoatomic ideal gas

Correct

Step-by-Step Solution

To determine the correct statements, let us recall the expressions for the molar heat capacities at constant volume (CvC_v) and constant pressure (CpC_p) for ideal gases in terms of the universal gas constant RR:

For an ideal gas with ff degrees of freedom:
Cv=f2RC_v = \frac{f}{2}R
Cp=Cv+R=(f2+1)RC_p = C_v + R = \left(\frac{f}{2} + 1\right)R

  1. For a monoatomic ideal gas (f=3f = 3):
    Cv=32RC_v = \frac{3}{2}R
    Cp=52RC_p = \frac{5}{2}R

  2. For a diatomic ideal gas (considering translational and rotational degrees of freedom, f=5f = 5):
    Cv=52RC_v = \frac{5}{2}R
    Cp=72RC_p = \frac{7}{2}R
    (Even if vibrational modes are active, f≥5f \geq 5, so both CvC_v and CpC_p are strictly greater for a diatomic gas).

Now, let us evaluate the given options:

  • Option (A):
    By Mayer's relation, for any ideal gas:
    Cp−Cv=RC_p - C_v = R
    This difference is constant and identical for both monoatomic and diatomic ideal gases.
    Thus, Option (A) is incorrect.

  • Option (B):

    • For a monoatomic gas:
      Cp+Cv=52R+32R=4RC_p + C_v = \frac{5}{2}R + \frac{3}{2}R = 4R
    • For a diatomic gas:
      Cp+Cv=72R+52R=6RC_p + C_v = \frac{7}{2}R + \frac{5}{2}R = 6R
      Since 6R>4R6R > 4R, Cp+CvC_p + C_v is larger for a diatomic ideal gas.
      Thus, Option (B) is correct.
  • Option (C):
    The adiabatic index γ=CpCv=1+2f\gamma = \frac{C_p}{C_v} = 1 + \frac{2}{f}:

    • For a monoatomic gas:
      CpCv=53≈1.67\frac{C_p}{C_v} = \frac{5}{3} \approx 1.67
    • For a diatomic gas:
      CpCv=75=1.40\frac{C_p}{C_v} = \frac{7}{5} = 1.40
      Since 75<53\frac{7}{5} < \frac{5}{3}, the ratio CpCv\frac{C_p}{C_v} is smaller for a diatomic ideal gas.
      Thus, Option (C) is incorrect.
  • Option (D):

    • For a monoatomic gas:
      Cp⋅Cv=(52R)(32R)=154R2=3.75R2C_p \cdot C_v = \left(\frac{5}{2}R\right)\left(\frac{3}{2}R\right) = \frac{15}{4}R^2 = 3.75R^2
    • For a diatomic gas:
      Cp⋅Cv=(72R)(52R)=354R2=8.75R2C_p \cdot C_v = \left(\frac{7}{2}R\right)\left(\frac{5}{2}R\right) = \frac{35}{4}R^2 = 8.75R^2
      Since 354R2>154R2\frac{35}{4}R^2 > \frac{15}{4}R^2, the product Cp⋅CvC_p \cdot C_v is larger for a diatomic ideal gas.
      Thus, Option (D) is correct.

Therefore, the correct options are B and D.