Question:
\(10 \text{ g}\) each of \(\text{CH}_4\) and \(\text{O}_2\) are kept in cylinders of same volume under same temperature, give the pressure ratio of two gases:
\(10 \text{ g}\) each of \(\text{CH}_4\) and \(\text{O}_2\) are kept in cylinders of same volume under same temperature, give the pressure ratio of two gases:
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For equal masses of two gases at the same temperature and volume, the pressure ratio is simply the inverse ratio of their molar masses: \(\frac{P_1}{P_2} = \frac{M_2}{M_1}\). Since oxygen is twice as heavy as methane, it has half as many particles, which means it exerts exactly half the pressure!
Updated On: May 25, 2026
- \(3 : 4\)
- \(2 : 3\)
- \(1 : 4\)
- \(2 : 1\)
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The Correct Option is D
Solution and Explanation
Concept:
According to the Ideal Gas Law, the pressure (\(P\)) exerted by a gas is related to its number of moles (\(n\)), absolute temperature (\(T\)), and volume (\(V\)) by the equation:
\[
PV = nRT \quad \Rightarrow \quad P = \frac{nRT}{V}
\]
When different gases are stored under identical conditions of volume (\(V\)) and temperature (\(T\)), the terms \(R\), \(T\), and \(V\) are constants. Therefore, the pressure exerted by a gas is directly proportional to its number of moles:
\[
P \propto n \quad \Rightarrow \quad \frac{P_{\text{CH}_4}}{P_{\text{O}_2}} = \frac{n_{\text{CH}_4}}{n_{\text{O}_2}}
\]
The number of moles is calculated using the formula:
\[
n = \frac{\text{Given Mass } (m)}{\text{Molar Mass } (M)}
\]
Step 1: Identify the molar masses of both gases.
• Molar mass of methane (\(\text{CH}_4\)): \[ M_{\text{CH}_4} = 12 + (4 \times 1) = 16 \text{ g mol}^{-1} \]
• Molar mass of oxygen gas (\(\text{O}_2\)): \[ M_{\text{O}_2} = 2 \times 16 = 32 \text{ g mol}^{-1} \]
Step 2: Calculate the number of moles for each gas.
We are given that the mass of both gases is exactly \(m = 10 \text{ g}\):
• Moles of \(\text{CH}_4\): \[ n_{\text{CH}_4} = \frac{10}{16} \text{ mol} \]
• Moles of \(\text{O}_2\): \[ n_{\text{O}_2} = \frac{10}{32} \text{ mol} \]
Step 3: Determine the ratio of their pressures.
Using the proportionality relationship \(\frac{P_{\text{CH}_4}}{P_{\text{O}_2}} = \frac{n_{\text{CH}_4}}{n_{\text{O}_2}}\): \[ \frac{P_{\text{CH}_4}}{P_{\text{O}_2}} = \frac{\frac{10}{16}}{\frac{10}{32}} \] Simplify the complex fraction by cancelling out the common mass factor of \(10\): \[ \frac{P_{\text{CH}_4}}{P_{\text{O}_2}} = \frac{32}{16} = \frac{2}{1} \] Thus, the pressure ratio of \(\text{CH}_4\) to \(\text{O}_2\) is exactly \(2 : 1\).
• Molar mass of methane (\(\text{CH}_4\)): \[ M_{\text{CH}_4} = 12 + (4 \times 1) = 16 \text{ g mol}^{-1} \]
• Molar mass of oxygen gas (\(\text{O}_2\)): \[ M_{\text{O}_2} = 2 \times 16 = 32 \text{ g mol}^{-1} \]
Step 2: Calculate the number of moles for each gas.
We are given that the mass of both gases is exactly \(m = 10 \text{ g}\):
• Moles of \(\text{CH}_4\): \[ n_{\text{CH}_4} = \frac{10}{16} \text{ mol} \]
• Moles of \(\text{O}_2\): \[ n_{\text{O}_2} = \frac{10}{32} \text{ mol} \]
Step 3: Determine the ratio of their pressures.
Using the proportionality relationship \(\frac{P_{\text{CH}_4}}{P_{\text{O}_2}} = \frac{n_{\text{CH}_4}}{n_{\text{O}_2}}\): \[ \frac{P_{\text{CH}_4}}{P_{\text{O}_2}} = \frac{\frac{10}{16}}{\frac{10}{32}} \] Simplify the complex fraction by cancelling out the common mass factor of \(10\): \[ \frac{P_{\text{CH}_4}}{P_{\text{O}_2}} = \frac{32}{16} = \frac{2}{1} \] Thus, the pressure ratio of \(\text{CH}_4\) to \(\text{O}_2\) is exactly \(2 : 1\).
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