Question

Two vessels A and B are of the same size and are at the same temperature. A contains 1 g of hydrogen and B contains 1 g of oxygen. \( P_A \) and \( P_B \) are the pressures of the gases in A and B respectively, then \( \frac{P_A}{P_B} \) is:

• A. 8
• B. 16
• C. 32
• D. 4

Hint:

The pressure of a gas is directly proportional to the number of moles of gas when volume and temperature are kept constant.

Answer 1

The Correct Option is B

Step 1: {Ideal Gas Equation}
The ideal gas equation is given by: \[ PV = nRT \] where:
\( P \) is the pressure of the gas
\( V \) is the volume of the gas
\( n \) is the number of moles of gas
\( R \) is the ideal gas constant
\( T \) is the temperature of the gas
Step 2: {Number of Moles of Hydrogen (\( n_H \))}
The molar mass of hydrogen (\( H_2 \)) is 2 g/mol. \[ n_H = \frac{{mass of hydrogen}}{{molar mass of hydrogen}} = \frac{1 { g}}{2 { g/mol}} = \frac{1}{2} { mol} \] Step 3: {Number of Moles of Oxygen (\( n_O \))}
The molar mass of oxygen (\( O_2 \)) is 32 g/mol. \[ n_O = \frac{{mass of oxygen}}{{molar mass of oxygen}} = \frac{1 { g}}{32 { g/mol}} = \frac{1}{32} { mol} \] Step 4: {Applying Ideal Gas Equation to both vessels}
Since the vessels are of the same size and at the same temperature, \( V \) and \( T \) are the same for both vessels. Therefore, we can write: For vessel A (hydrogen): \[ P_A V = n_H RT \] For vessel B (oxygen): \[ P_B V = n_O RT \] Step 5: {Finding the ratio \( \frac{P_A}{P_B} \)}
Divide the equation for vessel A by the equation for vessel B: \[ \frac{P_A V}{P_B V} = \frac{n_H RT}{n_O RT} \] \[ \frac{P_A}{P_B} = \frac{n_H}{n_O} = \frac{\frac{1}{2}}{\frac{1}{32}} = \frac{1}{2} \times \frac{32}{1} = 16 \] Thus, \( \frac{P_A}{P_B} = 16 \).