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Question Number 52992 by prakash jain last updated on 16/Jan/19

Commented by prakash jain last updated on 16/Jan/19

$$\mathrm{Correct}\mathrm{answer}\mathrm{is}\frac{4-\mathrm{R}}{2\mathrm{R}},\mathrm{NTA}\mathrm{has}\mathrm{given}$$$$\frac{4+\mathrm{R}}{2\mathrm{R}}.\mathrm{This}\mathrm{question}\mathrm{is}\mathrm{also}\mathrm{from}$$$$9\mathrm{th}\mathrm{Jan}\mathrm{morning}\mathrm{JE}E\mathrm{main}\mathrm{exam}.$$

Answered by tanmay.chaudhury50@gmail.com last updated on 16/Jan/19

$$P_A=N_A\left(\frac{RT}{V}\right)$$$$P_B=N_B\left(\frac{RT}{V}\right)$$$$P=P_A+P_B=\frac{RT}{V}(N_A+N_B)$$$$N_A+N_B=\frac{PV}{RT}$$$$N_B=\frac{PV}{RT}-N_A$$$$N_B=\frac{200\times 10}{R\times 1000}-0.5$$$$N_B=\frac{2}{R}-\frac{1}{2}$$$$N_B=\frac{4-R}{2R}$$

Answered by afachri last updated on 16/Jan/19

Answered by FelipeLz last updated on 26/Oct/21

$$PV=nRT$$$$2.{10}^2\cdot 10=(\frac{1}{2}+x)\cdot R\cdot {10}^3$$$$2=(\frac{1}{2}+x)\cdot R$$$$2=(\frac{1}{2}+x)\cdot R$$$$\frac{2}{R}=\frac{1}{2}+x$$$$x=\frac{2}{R}-\frac{1}{2}$$$$x=\frac{4-R}{2R}$$

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