Question:
Two identical cylinders contain helium at 2.5 atm and argon at 1 atm respectively. If both the gases are filled in one of the cylinders, the pressure would be:
Two identical cylinders contain helium at 2.5 atm and argon at 1 atm respectively. If both the gases are filled in one of the cylinders, the pressure would be:
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When gases from equal volumes at same temperature are combined into one of them, pressures simply add.
Updated On: Apr 18, 2026
- 3.5 atm
- 1.50 atm
- 1.75 atm
- 1 atm
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The Correct Option is A
Solution and Explanation
Concept:
Using ideal gas law:
\[
PV = nRT
\]
Step 1: Initial moles.
For identical cylinders (same \(V, T\)): \[ n_1 = \frac{P_1 V}{RT}, \quad n_2 = \frac{P_2 V}{RT} \]
Step 2: Total moles.
\[ n_{\text{total}} = \frac{(P_1 + P_2)V}{RT} \]
Step 3: Final pressure.
Now gases are in one cylinder (volume \(V\)): \[ P = \frac{n_{\text{total}} RT}{V} = P_1 + P_2 \] \[ P = 2.5 + 1 = 3.5 \, \text{atm} \]
Step 1: Initial moles.
For identical cylinders (same \(V, T\)): \[ n_1 = \frac{P_1 V}{RT}, \quad n_2 = \frac{P_2 V}{RT} \]
Step 2: Total moles.
\[ n_{\text{total}} = \frac{(P_1 + P_2)V}{RT} \]
Step 3: Final pressure.
Now gases are in one cylinder (volume \(V\)): \[ P = \frac{n_{\text{total}} RT}{V} = P_1 + P_2 \] \[ P = 2.5 + 1 = 3.5 \, \text{atm} \]
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