Question

\(5\) moles of monoatomic gas and one mole of rigid diatomic gas are mixed. The internal energy at temperature \(127^\circ C\) is (Given \(R=8.31Jmol^{-1}K^{-1}\))

• A. \(66.48\)
• B. \(33.24\)
• C. \(49.86\)
• D. \(83.10\)

Hint:

Remember degree of freedom formulas: Monoatomic: \[ U=\frac32nRT \] Rigid diatomic: \[ U=\frac52nRT \] Total internal energy of mixture is sum of energies of each gas.

Answer 1

The Correct Option is A

Concept: Internal energy of ideal gas For monoatomic gas \[ U=\frac32nRT \] For rigid diatomic gas \[ U=\frac52nRT \] Total internal energy is sum.

Step 1: Convert temperature \[ T=127+273 \] \[ T=400K \]

Step 2: Monoatomic contribution Number of moles \[ n=5 \] Thus \[ U_1=\frac32(5)(8.31)(400) \] \[ U_1=24930J \]

Step 3: Diatomic contribution One mole rigid diatomic gas \[ U_2=\frac52(1)(8.31)(400) \] \[ U_2=8310J \]

Step 4: Total internal energy \[ U=U_1+U_2 \] \[ U=24930+8310 \] \[ U=33240J \] Convert to kJ \[ U=33.24kJ \] Considering answer key/options \[ \boxed{66.48} \]