Question

Two closed vessels of same volume are joined through a narrow tube and both vessels are filled with air of pressure $90\text{ kPa}$ and temperature $400\text{ K}$. Keeping the temperature of one vessel constant at $400\text{ K}$ the second vessel temperature is raised to $500\text{ K}$. The final pressure in the vessels is _______ $\text{kPa}$.

• A. 100
• B. 120
• C. 90
• D. 105

Hint:

Calculate the total number of moles of gas initially using the ideal gas law $PV=nRT$. Since the system is closed, the total number of moles remains the same even after the temperature of one vessel is changed.

Answer 1

The Correct Option is A

We can solve this using the principle of conservation of the total number of moles of gas. Let $n_1$ and $n_2$ be the initial moles in the two vessels, and $n_1'$ and $n_2'$ be the moles after the temperature change.
Total moles initially: $n = \frac{P_1 V}{R T_1} + \frac{P_1 V}{R T_1} = \frac{2 P_1 V}{R T_1}$
Total moles finally: $n = \frac{P_f V}{R T_1} + \frac{P_f V}{R T_2}$
Since the total quantity of gas is constant, we set the two expressions equal:
$$ \frac{2 P_1 V}{R T_1} = \frac{P_f V}{R} \left( \frac{1}{T_1} + \frac{1}{T_2} \right) $$
Substituting the known values $P_1 = 90\text{ kPa}$, $T_1 = 400\text{ K}$, and $T_2 = 500\text{ K}$:
$$ \frac{2 \times 90}{400} = P_f \left( \frac{1}{400} + \frac{1}{500} \right) $$
$$ \frac{180}{400} = P_f \left( \frac{5 + 4}{2000} \right) $$
$$ 0.45 = P_f \left( \frac{9}{2000} \right) $$
$$ P_f = \frac{0.45 \times 2000}{9} = \frac{900}{9} = 100\text{ kPa} $$
Thus, the final common pressure in the two vessels is $100\text{ kPa}$, which is option 1.