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Standard X

Chemistry

Mole Concept

A closed vess...

Question

A closed vessel A having volume V contains Nitrogen $(N_{2})$ at Pressure P and temperature T. Another closed vessel B having the same volume V contains helium (He) at the same pressure P but temperature 2T (twice that in vessel A). Find The ratio of masses of $N_{2}$ and He in vessels A and B.

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Solution

The correct option is A
$F o r N_{2}, P = \frac{m_{N_{2}}}{M_{N_{2}}} \frac{R T}{V} (1) F o r H e, P = \frac{m_{(H e)}}{M_{(H e)}} \frac{(R) (2 t)}{V} (2) Equating (1) and (2) \frac{m_{N_{2}}}{25 \times 10^{- 3}} \frac{R T}{V} = \frac{m_{H e}}{4 \times 10^{- 3}} \frac{R T \times 2}{V} \frac{m_{N_{2}}}{m_{(H e)}} = \frac{28}{4} \times 2 = \frac{14}{1}$

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A closed vessel A having volume V contains Nitrogen (N2) at Pressure P and temperature T. Another closed vessel B having the same volume V contains helium (He) at the same pressure P but temperature 2T (twice that in vessel A). Find The ratio of masses of N2 and He in vessels A and B.

The correct option is A For N2,P=mN2MN2RTV (1)For He,P=m(He)M(He)(R)(2t)V (2)Equating (1) and (2)mN225×10−3RTV=mHe4×10−3RT×2VmN2m(He)=284×2=141

A closed vessel A having volume V contains Nitrogen $(N_{2})$ at Pressure P and temperature T. Another closed vessel B having the same volume V contains helium (He) at the same pressure P but temperature 2T (twice that in vessel A). Find The ratio of masses of $N_{2}$ and He in vessels A and B.