Question

A and B are two ideal gases. \(3\) g-mole of gas A at absolute temperature \(T_1\) and \(5\) g-mole of gas B at absolute temperature \(T_2\) have been mixed. There is no loss of energy in the process. Find the temperature of the mixture if \(T_1=300\) K and \(T_2=500\) K.

• A. \(350\) K
• B. \(401.5\) K
• C. \(425\) K
• D. \(450\) K

Hint:

For adiabatic mixing of ideal gases, \[ T_{\text{final}} = \frac{\sum n_iT_i}{\sum n_i} \] when the molar heat capacities are the same.

Answer 1

The Correct Option is C

Concept: Since there is no heat loss, \[ \text{Heat lost}=\text{Heat gained} \] For ideal gases, internal energy is proportional to the number of moles and temperature. Hence, \[ n_1C_V(T-T_1)+n_2C_V(T-T_2)=0 \] Since both gases are ideal and \(C_V\) cancels, \[ T=\frac{n_1T_1+n_2T_2}{n_1+n_2}. \]

Step 1: Substitute the given values. \[ n_1=3,\qquad T_1=300\text{ K} \] \[ n_2=5,\qquad T_2=500\text{ K} \] \[ T = \frac{3(300)+5(500)}{3+5} \] \[ = \frac{900+2500}{8} \] \[ = \frac{3400}{8} \] \[ = 425\text{ K} \]

Step 2: Write the final answer. \[\begin{aligned} \boxed{425\text{ K}} \end{aligned}\] Hence, option \(\mathbf{(C)}\) is correct.