Question

A closed vessel contains 0.1 mole of a monoatomic ideal gas at 200 K . If 0.05 mole of the same gas at 400 K is added to it, the final equilibrium temperature (in K ) of the gas in the vessel will be close to ______

Answer 1

267

Answer 2

The problem involves mixing two samples of the same monoatomic ideal gas at different temperatures in a closed vessel. Assuming no heat exchange with the surroundings, the total internal energy of the system remains conserved.

For an ideal gas, the internal energy $U$ is given by: $U = nC_vT$ For a monoatomic ideal gas, the molar heat capacity at constant volume is $C_v = \frac{3}{2}R$. So, the internal energy can be written as: $U = n \left(\frac{3}{2}R\right) T$

Let's denote the initial state of the first gas sample as (1) and the second gas sample as (2).

For gas 1: Number of moles, $n_1 = 0.1$ mol Initial temperature, $T_1 = 200$ K Initial internal energy, $U_1 = n_1 \left(\frac{3}{2}R\right) T_1$

For gas 2: Number of moles, $n_2 = 0.05$ mol Initial temperature, $T_2 = 400$ K Initial internal energy, $U_2 = n_2 \left(\frac{3}{2}R\right) T_2$

When the two gases are mixed, they reach a final equilibrium temperature, $T_f$. The total number of moles in the final mixture is $n_{total} = n_1 + n_2$. The final internal energy of the mixture is $U_{final} = (n_1 + n_2) \left(\frac{3}{2}R\right) T_f$.

According to the conservation of internal energy: $U_{initial} = U_{final}$ $U_1 + U_2 = U_{final}$ $n_1 \left(\frac{3}{2}R\right) T_1 + n_2 \left(\frac{3}{2}R\right) T_2 = (n_1 + n_2) \left(\frac{3}{2}R\right) T_f$

We can cancel the common term $\left(\frac{3}{2}R\right)$ from both sides: $n_1 T_1 + n_2 T_2 = (n_1 + n_2) T_f$

Now, solve for $T_f$: $T_f = \frac{n_1 T_1 + n_2 T_2}{n_1 + n_2}$

Substitute the given values: $T_f = \frac{(0.1 \text{ mol} \times 200 \text{ K}) + (0.05 \text{ mol} \times 400 \text{ K})}{0.1 \text{ mol} + 0.05 \text{ mol}}$ $T_f = \frac{20 \text{ K} + 20 \text{ K}}{0.15 \text{ mol}}$ $T_f = \frac{40}{0.15}$ $T_f = \frac{4000}{15}$ $T_f = \frac{800}{3}$ $T_f \approx 266.67 \text{ K}$

Rounding to the nearest integer, the final equilibrium temperature is approximately 267 K.