Question

At T (K), equal weights of H$_2$, D$_2$, and T$_2$ are present in closed vessel. The pressure exerted by this gaseous mixture is P atm. The ratio of partial pressures of T$_2$, D$_2$ and H$_2$ is approximately (H, D and T are isotopes of hydrogen).

• A. 0.33 : 0.33 : 0.33
• B. 0.18 : 0.27 : 0.54
• C. 0.25 : 0.50 : 0.25
• D. 0.54 : 0.27 : 0.18

Hint:

The partial pressure of a gas is proportional to the number of moles, and for equal masses of different gases, it is inversely proportional to their molar masses.

Answer 1

The Correct Option is B

At constant temperature, the pressure exerted by a gas in a closed vessel is proportional to the number of molecules present. Since equal weights of H$_2$, D$_2$, and T$_2$ are given, the ratio of partial pressures depends on the molar masses. The ratio of partial pressures of gases with the same volume is inversely proportional to their molar masses. Thus, for equal masses: \[ \frac{P_{T_2}}{P_{H_2}} = \frac{m_{H_2}}{m_{T_2}} \quad \text{and similarly for D$_2$ and H$_2$} \] Given that the molar masses of H$_2$, D$_2$, and T$_2$ are approximately in the ratio 1 : 2 : 3, the partial pressure ratio will be: \[ P_{T_2} : P_{D_2} : P_{H_2} = 0.18 : 0.27 : 0.54 \]

Answer 2

Step 1: Understanding the Problem
Equal weights of hydrogen isotopes H\(_2\), D\(_2\), and T\(_2\) are in a closed vessel at temperature T. We need to find the ratio of their partial pressures.

Step 2: Molar Masses of the Gases
- Molar mass of H\(_2\) = 2 g/mol
- Molar mass of D\(_2\) = 4 g/mol
- Molar mass of T\(_2\) = 6 g/mol

Step 3: Calculate Number of Moles for Each Gas
Since equal masses \(m\) of each gas are taken,
\[ n = \frac{m}{M} \]
Let the equal mass be \(m\) grams.
Then,
Number of moles of H\(_2\), \(n_H = \frac{m}{2}\)
Number of moles of D\(_2\), \(n_D = \frac{m}{4}\)
Number of moles of T\(_2\), \(n_T = \frac{m}{6}\)

Step 4: Relationship Between Partial Pressure and Moles
At the same temperature and volume,
\[ P_i \propto n_i \]
So, partial pressures ratio is:
\[ P_T : P_D : P_H = n_T : n_D : n_H = \frac{m}{6} : \frac{m}{4} : \frac{m}{2} = \frac{1}{6} : \frac{1}{4} : \frac{1}{2} \]
To simplify, multiply each term by 12 (LCM of 6, 4, 2):
\[ 2 : 3 : 6 \]
Divide all by 11 to get approximate decimal ratio:
\[ 0.18 : 0.27 : 0.54 \]

Step 5: Conclusion
The ratio of partial pressures of T\(_2\), D\(_2\), and H\(_2\) is approximately 0.18 : 0.27 : 0.54.