Question

At 300 K, 1 g of an ideal gas (A) exerted a pressure of 2 atm in a closed container. When 2 g of another ideal gas (B) was introduced into this container at the same temperature, the pressure changed to 3 atm. The correct relationship between the molar mass of A (\(M_A\)) and B (\(M_B\)) is:

• A. \( 8M_A = M_B \)
• B. \( M_A = 4M_B \)
• C. \( M_A = M_B \)
• D. \( 4M_A = M_B \)

Hint:

When two ideal gases are mixed, the total pressure is proportional to the sum of the number of moles, and the number of moles is inversely proportional to the molar mass.

Answer 1

The Correct Option is D

From the ideal gas law, we know that the pressure is directly proportional to the number of moles and inversely proportional to the molar mass: \[ P \propto \frac{n}{M} \] Let the molar mass of gas \( A \) be \( M_A \) and that of gas \( B \) be \( M_B \). The pressure exerted by gas \( A \) is 2 atm, and the total pressure after introducing gas \( B \) is 3 atm. From the ideal gas law: \[ P_{\text{total}} \propto \frac{n_A + n_B}{M_A + M_B} \] After solving for the relationship between the molar masses, we get: \[ M_A = 4 M_B \] Thus, the correct answer is Option 4: \( 4M_A = M_B \).