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# Please tell above answer quickly and as soon as possible   ........................

Samyak Jain
333 Points
2 years ago
RMS speed of ideal gas molecules is $\dpi{100} \sqrt{3RT/M}$ , where T is the temperature in kelvin, R is universal gas constant,
M is the molar mass of gas.
$\dpi{100} \therefore$ ratio of rms speeds of molecules at T2 and T1 is $\dpi{80} \sqrt{3RT_2 / M} / \sqrt{3RT_1 / M}$ = $\dpi{80} \sqrt{T_2 / T_1}$  …..(1)
Now, we need to compute temperatures T1 and T2.
We know that for isotherm, PV = constant i.e. PV = nRT and T is constant.
Consider isotherm at T1. It is passing through P = 1.105 Pa and V = 1 m3.
So, 1.105 Pa . 1 m3 = nR T1  or  T1 = (105 / nR) K
Similarly, for T2, P = 2.105 Pa and V = 2 m3.
So, 2.105 Pa . 2 m3 = nR T2  or  T2 = (4.105 / nR) K
Substitute value of T1 and T2 in (1) and you will get
ratio of rms speeds = $\dpi{80} \sqrt{(4.10^5 / nR) K / (1.10^5 / nR) K}$   =  2 .