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 _______ $\text{kPa}$.
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 _______ $\text{kPa}$.
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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.
Updated On: Apr 9, 2026
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The Correct Option is A
Solution and Explanation
This problem involves the application of the Ideal Gas Law, $PV = nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the universal gas constant, and $T$ is temperature. Initially, we have two vessels of volume $V$, each at pressure $P_i = 90\text{ kPa}$ and temperature $T_i = 400\text{ K}$. The total number of moles of air in the system is:
$$ n_{total} = n_1 + n_2 = \frac{P_i V}{R T_i} + \frac{P_i V}{R T_i} = \frac{2 P_i V}{R T_i} $$
Substituting the initial values:
$$ n_{total} = \frac{2 \times 90 \times V}{R \times 400} = \frac{180 V}{400 R} = \frac{9 V}{20 R} $$
In the final state, the temperature of vessel 1 remains $T_1 = 400\text{ K}$ and the temperature of vessel 2 is raised to $T_2 = 500\text{ K}$. Let the final common pressure in both vessels be $P_f$. Since the system is closed, the total number of moles remains constant:
$$ n_{total} = \frac{P_f V}{R T_1} + \frac{P_f V}{R T_2} = \frac{P_f V}{R} \left( \frac{1}{400} + \frac{1}{500} \right) $$
Equating the expressions for $n_{total}$ from the initial and final states:
$$ \frac{9 V}{20 R} = \frac{P_f V}{R} \left( \frac{5 + 4}{2000} \right) = \frac{P_f V}{R} \cdot \frac{9}{2000} $$
Simplifying the equation by canceling $V/R$ and the factor of $9$:
$$ \frac{1}{20} = \frac{P_f}{2000} $$
$$ P_f = \frac{2000}{20} = 100\text{ kPa} $$
The final pressure in the vessels is $100\text{ kPa}$.
$$ n_{total} = n_1 + n_2 = \frac{P_i V}{R T_i} + \frac{P_i V}{R T_i} = \frac{2 P_i V}{R T_i} $$
Substituting the initial values:
$$ n_{total} = \frac{2 \times 90 \times V}{R \times 400} = \frac{180 V}{400 R} = \frac{9 V}{20 R} $$
In the final state, the temperature of vessel 1 remains $T_1 = 400\text{ K}$ and the temperature of vessel 2 is raised to $T_2 = 500\text{ K}$. Let the final common pressure in both vessels be $P_f$. Since the system is closed, the total number of moles remains constant:
$$ n_{total} = \frac{P_f V}{R T_1} + \frac{P_f V}{R T_2} = \frac{P_f V}{R} \left( \frac{1}{400} + \frac{1}{500} \right) $$
Equating the expressions for $n_{total}$ from the initial and final states:
$$ \frac{9 V}{20 R} = \frac{P_f V}{R} \left( \frac{5 + 4}{2000} \right) = \frac{P_f V}{R} \cdot \frac{9}{2000} $$
Simplifying the equation by canceling $V/R$ and the factor of $9$:
$$ \frac{1}{20} = \frac{P_f}{2000} $$
$$ P_f = \frac{2000}{20} = 100\text{ kPa} $$
The final pressure in the vessels is $100\text{ kPa}$.
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