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        Please tell above answer quickly and as soon as possible   ........................
5 months ago

Samyak Jain
328 Points
							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) KSimilarly, for T2, P = 2.105 Pa and V = 2 m3.So, 2.105 Pa . 2 m3 = nR T2  or  T2 = (4.105 / nR) KSubstitute value of T1 and T2 in (1) and you will getratio of rms speeds = $\dpi{80} \sqrt{(4.10^5 / nR) K / (1.10^5 / nR) K}$   =  2 .

5 months ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions