Question-52992

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

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

Correct answer is \frac{4 - R}{2 R}, NTA has given

\frac{4 + R}{2 R} . This question is also from

9 th Jan morning JE E main exam .

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

P_{A} = N_{A} (\frac{R T}{V})

P_{B} = N_{B} (\frac{R T}{V})

P = P_{A} + P_{B} = \frac{R T}{V} (N_{A} + N_{B})

N_{A} + N_{B} = \frac{P V}{R T}

N_{B} = \frac{P V}{R T} - 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}{2 R}

Answered by afachri last updated on 16/Jan/19

Answered by FelipeLz last updated on 26/Oct/21

P V = n R T

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}{2 R}

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