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

 

To find the ratio of the root mean square (rms) velocities of gas $X$ and gas $Y$ at 300 K, we'll follow these steps:

Step 1: Understand the given data

• Gas $X$:

  • Mass ($m_X$) = 10 g

  • Pressure ($P_X$) = 2 atm

  • Temperature ($T$) = 300 K

• Gas $Y$:

  • Mass ($m_Y$) = 80 g

  • When added to the vessel, the total pressure becomes 6 atm.

Step 2: Determine the molar masses

Let the molar masses of $X$ and $Y$ be $M_X$ and $M_Y$, respectively.

For Gas $X$:

Using the ideal gas law:

$$P_{X}V=n_{X}RT$$

where $n_X=\frac{m_X}{M_X}=\frac{10}{M_X}$.

So,

$$2V=\left(\frac{10}{M_X}\right)R\times 300$$$$V=\frac{10\times R\times 300}{2M_X}=\frac{1500R}{M_X}$$

For Gas $X$ and $Y$ together:

Total pressure $P_{\text{total}}=6$ atm.

Total moles:

$$n_{\text{total}}=n_X+n_Y=\frac{10}{M_X}+\frac{80}{M_Y}$$

Using the ideal gas law again:

$$6V=(\frac{10}{M_X}+\frac{80}{M_Y})R\times 300$$

Substitute $V$ from the first equation:

$$6\left(\frac{1500R}{M_X}\right)=(\frac{10}{M_X}+\frac{80}{M_Y})R\times 300$$

Simplify:

$$\frac{9000R}{M_X}=(\frac{10}{M_X}+\frac{80}{M_Y})300R$$$$\frac{9000}{M_X}=\frac{3000}{M_X}+\frac{24000}{M_Y}$$$$\frac{6000}{M_X}=\frac{24000}{M_Y}$$$$\frac{M_Y}{M_X}=\frac{24000}{6000}=4$$$$M_Y=4M_X$$

Step 3: Calculate the rms velocities

The root mean square velocity is given by:

$$v_{\text{rms}}=\sqrt{\frac{3RT}{M}}$$

The ratio of the rms velocities of $X$ and $Y$ is:

$$\frac{v_{\text{rms},X}}{v_{\text{rms},Y}}=\sqrt{\frac{M_Y}{M_X}}=\sqrt{4}=2$$

Final Answer:

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