Question

Two thermally insulated vessels $1$ and $2$ are filled with air at temperature ($T_1$, $T_2$), volume ($V_1$, $V_2$) and pressure ($P_1$, $P_2$) respectively. If the value joining the two vessels is opened, the temperature inside the vessel at equilibrium will be

• A. T₁ +T₂
• B. (T₁ + T₂)/2
• C. T₁T₂(P₁V₁+P₂V₂)/P₁V₁T₂+P₂V₂T₁)
• D. T₁T₂(P₁V₁+P₂V₂)/P₁VT₁+P₂V₂T₂)

Answer 1

The correct option is C

T₁T₂(P₁V₁+P₂V₂)/P₁V₁T₂+P₂V₂T₁)

According to the kinetic theory, the average kinetic energy(KE) per molecule of a gas

= $\frac{3}{2}KT$. Let $n_1$ and $n_2$ be the number of moles of air in vessels 1 and 2 respectively.

Before mixing, the total KE of molecules in the two vessels is

$E_1=\frac{3}{2}{n}_{1}k{T}_1+\frac{3}{2}{n}_{2}k{T}_2$

= $\frac{3}{2}k(n_1{T}_1+n_2{T}_2)$

After mixing, the total KE of molecules is

$E_2=\frac{3}{2}(n_1+n_2)kT$

Where T is the temperature when equilibrium is established. Since there is no loss of energy

(Because the vessels are insulated), $E_2=E_1$ or

$\frac{3}{2}(n_1+n_2)kT=\frac{3}{2}k(n_1{T}_1+n_2{T}_2)$

or $T=\frac{n_1{T}_1+n_2{T}_2}{n_1+n_2}$

Now $P_1{V}_1=n_{1}R{T}_1$ and $P_2{V}_2=n_{2}R{T}_2$ which give

$n_1=\frac{P_1{V}_1}{R{T}_1}$ and $n_2=\frac{P_2{V}_2}{R{T}_2}$

Using these in Eq. (1) and simplifying, we get

$T=\frac{T_1{T}_2(P_1{V}_1+P_2{V}_2)}{(P_1{V}_1{T}_2+P_2{V}_2{T}_1)}$