Question

If \(2\) mole of an ideal monoatomic gas at temperature \(T\) is mixed with \(6\) mole of another ideal monoatomic gas at temperature \(2T\) then the temperature of mixture is :

• A. \( \frac{5}{2}T \)
• B. \( \frac{5}{4}T \)
• C. \( \frac{7}{2}T \)
• D. \( \frac{7}{4}T \)

Answer 1

The Correct Option is B

Concept: For mixing of ideal gases without heat loss: \[ \text{Total internal energy conserved} \] For monoatomic gas: \[ U=\frac{3}{2}nRT \] Step 1: {Write initial internal energies.} Gas \(1\): \[ U_1=\frac{3}{2}(2)RT=3RT \] Gas \(2\): \[ U_2=\frac{3}{2}(6)R(2T) \] \[ =18RT \] Step 2: {Total initial energy.} \[ U_{total}=21RT \] Step 3: {Final energy.} Total moles: \[ n=8 \] Final internal energy: \[ U=\frac{3}{2}(8)RT_f \] \[ =12RT_f \] Step 4: {Equate energies.} \[ 12RT_f=21RT \] \[ T_f=\frac{7}{4}T \] Thus the final temperature is \[ \frac{7}{4}T \]