Question:

A vessel contains \(3\) moles of He, \(1\) mole of Ar, \(5\) moles of \(N_2\), and \(3\) moles of \(H_2\). If the vibrational modes are ignored, the total internal energy of the system of gases is

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For ideal gases, \[ U=\frac{f}{2}nRT. \] Monoatomic gases have \[ f=3, \] while diatomic gases without vibrational modes have \[ f=5. \]
Updated On: Jun 26, 2026
  • \(20RT\)
  • \(26RT\)
  • \(25RT\)
  • \(30RT\)
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The Correct Option is B

Solution and Explanation

Step 1: Use the formula for internal energy of an ideal gas.
For an ideal gas, \[ U=\frac{f}{2}nRT, \] where \(f\) is the number of degrees of freedom.

Step 2: Internal energy of monoatomic gases.
He and Ar are monoatomic gases.
For a monoatomic gas, \[ f=3. \] So, \[ U_{\text{monoatomic}}=\frac{3}{2}nRT. \] Total moles of monoatomic gases: \[ n=3+1=4. \] Therefore, \[ U_{\text{He+Ar}}=\frac{3}{2}(4)RT. \] \[ U_{\text{He+Ar}}=6RT. \]

Step 3: Internal energy of diatomic gases.
\(N_2\) and \(H_2\) are diatomic gases.
Since vibrational modes are ignored, for a diatomic gas, \[ f=5. \] So, \[ U_{\text{diatomic}}=\frac{5}{2}nRT. \] Total moles of diatomic gases: \[ n=5+3=8. \] Therefore, \[ U_{N_2+H_2}=\frac{5}{2}(8)RT. \] \[ U_{N_2+H_2}=20RT. \]

Step 4: Find total internal energy.
\[ U_{\text{total}}=U_{\text{He+Ar}}+U_{N_2+H_2}. \] \[ U_{\text{total}}=6RT+20RT. \] \[ U_{\text{total}}=26RT. \]

Step 5: Final conclusion.
Therefore, the total internal energy of the system is \[ \boxed{26RT} \] Hence, the correct option is \[ \boxed{(2)} \]
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