Question

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

Answer 1

Step1: Given data and Assumption.

Number of mole of ideal gas, $n_1=0.01$

Temperature $T_1=200K$

Number of mole of ideal gas, $n_2=0.05$

Temperature $T_2=400K$

And work done on gas and heat supplied to the gas are zero.

Step2: Finding the final equilibrium temperature of the gas.

$U=n{C}_{v}T$

Where $U=$internal energy of the gas, $C_v=$specific heat constant, $T=$ Temperature, $n=$number of mole of gas.

Step 3: According to the question:

The reaction is under equilibrium. So, the initial internal energy is equal to the final internal energy of the gas.

$U_1+U_2=U_1\text{'}+U_2\text{'}$

$\Rightarrow$ $n_1{C}_v{T}_1+n_2{C}_v{T}_2=\left(n_1+n_2\right)C_{v}T$

$\Rightarrow$$\left(0.1\times C_v\times 200\right)+\left(0.05\times C_v\times 400\right)=\left(0.1+0.05\right)\times C_v\times T$

$\Rightarrow$ $20C_v+20C_v=0.15\times C_v\times T$

$\Rightarrow$ $T=\frac{40}{0.15}$

$\Rightarrow$ $T=266.67K$

Hence, the final equilibrium temperature $\left(inK\right)$ of the gas in the vessel will be close to $\mathbf{266}\mathbf{.}\mathbf{67}\mathit{K}$.