Question

A closed vessel contains 10 g of an ideal gas X at 300 K, which exerts 2 atm pressure. At the same temperature, 80 g of another ideal gas Y is added to it and the pressure becomes 6 atm. The ratio of root mean square velocities of X and Y at 300 K is

• A. 2√2:√3
• B. 2√2:1
• C. 1:2
• D. 2:1

Answer 1

The Correct Option is D

To solve the problem of finding the ratio of root mean square velocities of two ideal gases X and Y, we use the root mean square (RMS) velocity formula: \( v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \) where \( R \) is the gas constant, \( T \) is the temperature, and \( M \) is the molar mass of the gas. At a constant temperature \( T = 300 \, \text{K} \), the ratio of the RMS velocities for gases X and Y can be calculated as:

\[ \frac{v_{\text{rms,X}}}{v_{\text{rms,Y}}} = \sqrt{\frac{M_Y}{M_X}} \]

From the ideal gas law \( PV = nRT \), we know that at constant temperature and volume and for the same gas constant \( R \), the pressure is proportional to moles \( n \). Initially, 10 g of gas X exerts a pressure of 2 atm. When 80 g of gas Y is added, the pressure becomes 6 atm.

The additional moles contributed by gas Y increase the total pressure from 2 atm to 6 atm, thus the pressure due to gas Y alone is 4 atm. The pressure is directly proportional to the number of moles, so the moles of gas X and Y can be written as:

\[ \frac{n_Y}{n_X} = \frac{4}{2} = 2 \]

Using the ideal gas law, number of moles can be expressed as \( n = \frac{\text{mass}}{\text{molar mass}} \). Hence:

\[ \frac{\frac{80}{M_Y}}{\frac{10}{M_X}} = 2 \]

\[ \frac{80 M_X}{10 M_Y} = 2 \]

\[ \frac{8 M_X}{M_Y} = 2 \]

\[ 4 M_X = M_Y \]

Thus, \( M_Y = 4M_X \).

The ratio of the RMS velocities is:

\[ \frac{v_{\text{rms,X}}}{v_{\text{rms,Y}}} = \sqrt{\frac{4M_X}{M_X}} = \sqrt{4} = 2 \]

Therefore, the ratio of root mean square velocities \( v_{\text{rms,X}} : v_{\text{rms,Y}} \) is \( 2:1 \).

Correct answer: 2:1

Answer 2

To solve the problem, we need to find the ratio of the root mean square (RMS) velocities of two ideal gases X and Y at the same temperature.

1. Understanding RMS velocity:
The RMS velocity of a gas is given by:
\[ v_{\text{rms}} = \sqrt{\frac{3RT}{M}} \] where \(R\) is the gas constant, \(T\) is the temperature (in Kelvin), and \(M\) is the molar mass of the gas.

2. Given Data:
- Gas X: mass \(m_X = 10\, \text{g}\), pressure \(P_X = 2\, \text{atm}\)
- Gas Y: mass \(m_Y = 80\, \text{g}\), final total pressure \(P_{\text{total}} = 6\, \text{atm}\) after adding Y
- Temperature for both gases \(T = 300\, K\)

3. Finding moles of gases:
Let molar masses of gases X and Y be \(M_X\) and \(M_Y\) respectively.
Number of moles of X:
\[ n_X = \frac{m_X}{M_X} \] Number of moles of Y:
\[ n_Y = \frac{m_Y}{M_Y} \]

4. Using ideal gas law relation for pressure:
Pressure is proportional to the number of moles (at constant volume and temperature):
\[ \frac{P_Y}{P_X} = \frac{n_Y}{n_X} = \frac{80/M_Y}{10/M_X} = \frac{8 M_X}{M_Y} \] Given \(P_X = 2\, \text{atm}\), total pressure after adding Y is 6 atm, so:
\[ P_Y = 6 - 2 = 4\, \text{atm} \] Therefore, \[ \frac{P_Y}{P_X} = \frac{4}{2} = 2 = \frac{8 M_X}{M_Y} \Rightarrow \frac{M_Y}{M_X} = 4 \]

5. Calculating ratio of RMS velocities:
\[ \frac{v_{\text{rms},X}}{v_{\text{rms},Y}} = \sqrt{\frac{M_Y}{M_X}} = \sqrt{4} = 2 \] Hence, \[ \frac{v_{\text{rms},X}}{v_{\text{rms},Y}} = 2 : 1 \]

Final Answer:
The ratio of RMS velocities of X to Y is \(2 : 1\).