© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the given data
• The initial temperature is 27°C, which we convert to Kelvin:
T_{1} = 27 + 273 = 300\text{ K} .
• The initial pressure is 1\text{ atm} . (Note: RMS speed does not depend on external pressure for a fixed mass of gas.)
• The initial root mean square (rms) speed is v_{1} = 200\,\text{m s}^{-1} .
• The final temperature is 127°C, which we convert to Kelvin:
T_{2} = 127 + 273 = 400\text{ K} .
• The final pressure given is 2\text{ atm} , but that does not affect rms speed calculation for the same mass of gas.
• The final rms speed is given in the form
\frac{x}{\sqrt{3}}\, \text{m s}^{-1} .
Step 2: Recall the formula for RMS speed
For a gas of fixed mass m at temperature T , the rms speed v_{\mathrm{rms}} is:
v_{\mathrm{rms}} = \sqrt{\frac{3\,R\,T}{m}},
where R is the universal gas constant and m is the mass of one mole (or any fixed mass) of the gas.
Since R and m are constants for the same gas, v_{\mathrm{rms}} depends only on T .
Step 3: Write the relationship between the two rms speeds
For the same gas of the same mass at two different temperatures T_1 and T_2 , the ratio of rms speeds is:
\frac{v_{1}}{v_{2}} = \sqrt{\frac{T_1}{T_2}}.
Step 4: Substitute known values and solve for v_{2}
Using the given temperatures:
\frac{v_{1}}{v_{2}} = \sqrt{\frac{300}{400}} = \sqrt{\frac{3}{4}}.
Therefore,
v_{2} = v_{1}\,\sqrt{\frac{4}{3}} = 200 \times \sqrt{\frac{4}{3}} = 200 \times \frac{2}{\sqrt{3}} = \frac{400}{\sqrt{3}}\,\text{m s}^{-1}.
Step 5: Compare with the given form and find x
We are given that the final speed is
v_{2} = \frac{x}{\sqrt{3}}\,\text{m s}^{-1}.
We found
v_{2} = \frac{400}{\sqrt{3}}\,\text{m s}^{-1}.
Hence, matching the form
\frac{x}{\sqrt{3}} = \frac{400}{\sqrt{3}} ,
we deduce
x = 400.
Final Answer
\displaystyle 400
