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2 g of Hydrogen is mixed with 22.4 litre of Helium at STP in a container of volume 20 litre. If the final temperature is 300 K, find the pressure.

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$\begin{array}{rcl} 8.31 \times 10^{4} N / m^{2} \end{array}$

$\begin{array}{rcl} 24.93 \times 10^{4} N / m^{2} \end{array}$

$\begin{array}{rcl} 3 \times 10^{4} N / m^{2} \end{array}$

$\begin{array}{rcl} 24.93 N / m^{2} \end{array}$

answer is B.

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Detailed Solution

2 g of Hydrogen = 1 mole Hydrogen

22.4 litre He at STP = 1 mole of He

$\begin{array}{rcl} P & = & P_{H} + P_{He} \\ = & (n_{H} + n_{He}) \frac{RT}{V} \\ = & (1 + 1) \frac{8.31 \times 300}{20 \times 10^{- 3}} \\ = & 24.93 \times 10^{4} N / m^{2} \end{array}$

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