Question:

When an ideal diatomic gas is heated at constant pressure, the fraction of the heat utilised to increase the internal energy of the gas is

Show Hint

- Change in internal energy: \( \Delta U = nC_V \Delta T \). - Heat supplied at constant pressure: \( Q_P = nC_P \Delta T \). - Heat supplied at constant volume: \( Q_V = nC_V \Delta T = \Delta U \). - Fraction of heat for internal energy at constant pressure: \( \frac{\Delta U}{Q_P} = \frac{nC_V\Delta T}{nC_P\Delta T} = \frac{C_V}{C_P} = \frac{1}{\gamma} \). - For a diatomic gas, degrees of freedom \(f=5\). \( C_V = \frac{f}{2}R = \frac{5}{2}R \). \( C_P = C_V+R = \frac{7}{2}R \). \( \gamma = \frac{C_P}{C_V} = \frac{7}{5} \).
Updated On: Jun 5, 2025
  • \( \frac{2}{5} \)
  • \( \frac{3}{5} \)
  • \( \frac{3}{7} \)
  • \( \frac{5}{7} \)
Hide Solution
collegedunia
Verified By Collegedunia

The Correct Option is D

Solution and Explanation

For an ideal gas, the change in internal energy \( \Delta U \) is given by \( \Delta U = nC_V \Delta T \), where \(n\) is the number of moles, \(C_V\) is the molar specific heat at constant volume, and \( \Delta T \) is the change in temperature.
When heat \( Q \) is supplied at constant pressure, it is given by \( Q = nC_P \Delta T \), where \(C_P\) is the molar specific heat at constant pressure.
The fraction of heat utilised to increase the internal energy is \( \frac{\Delta U}{Q} \).
\[ \frac{\Delta U}{Q} = \frac{nC_V \Delta T}{nC_P \Delta T} = \frac{C_V}{C_P} \] The ratio \( \gamma = \frac{C_P}{C_V} \) is the adiabatic index.
So the fraction is \( \frac{1}{\gamma} \).
For a diatomic gas, the number of degrees of freedom (f) is typically 5 (3 translational + 2 rotational, neglecting vibrational modes at ordinary temperatures).
\( C_V = \frac{f}{2}R \).
For a diatomic gas, \( f=5 \), so \( C_V = \frac{5}{2}R \).
\( C_P = C_V + R = \frac{5}{2}R + R = \frac{7}{2}R \).
The adiabatic index \( \gamma = \frac{C_P}{C_V} = \frac{(7/2)R}{(5/2)R} = \frac{7}{5} \).
The fraction of heat utilised to increase internal energy is \( \frac{C_V}{C_P} = \frac{1}{\gamma} = \frac{1}{7/5} = \frac{5}{7} \).
This matches option (4).
Was this answer helpful?
0
0