Questions

Two vessels separately contain two ideal gases A and B at the same temperature, the pressure of A being twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio of molecular weight of A and B is

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a

2

b

12

c

23

d

34

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detailed solution

Correct option is D

According to an ideal gas equation, the molecular weight of an ideal gas isM=ρRTP as P=ρRTMwhere P, T and ρ are the pressure, temperature and density of the gas respectively and R is the universal gas constant.∴ The molecular weight of A isMA=ρARTAPAand that of B isMB=ρBRTBPB Hence, their corresponding ratio is MAMB=ρAρBTATBPBPA Here, ρAρB=1.5=32,TATB=1 and PAPB=2∴ MAMB=32(1)12=34

Similar Questions

For one-mole ideal gas, a process follows the relation $\mathrm{P}=\left(\frac{{\mathrm{V}}^{2}}{{\mathrm{V}}^{2}+{\mathrm{V}}_{0}^{2}}\right)\left({\mathrm{P}}_{0}\right),$ where P_{0} and V_{0} are constants. If the volume the gas changes from V_{0} to 2V_{0}. What is the change in temperature?

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