A vessel contains 0.5 g of hydrogen and 0.8 g of oxygen. The volume of vessel is 528 m3 while its temperature is maintained at 300 K. Then the pressure of the mixture is MH2=2, MO2=32, R=253J−mol−1K−1.

By Dalton’s law,P=PH2+PO2 =nH2RTV+nO2RTV=0.52+0.832253300528=3850Nm−2

A vessel contains 0.5 g of hydrogen and 0.8 g of oxygen. The volume of vessel is 528 m3 while its temperature is maintained at 300 K. Then the pressure of the mixture is MH2=2, MO2=32, R=253J−mol−1K−1.

By Dalton’s law,P=PH2+PO2 =nH2RTV+nO2RTV=0.52+0.832253300528=3850Nm−2

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A vessel contains 0.5 g of hydrogen and 0.8 g of oxygen. The volume of vessel is $\frac{5}{28} m^{3}$ while its temperature is maintained at 300 K. Then the pressure of the mixture is $(M_{H_{2}} = 2, M_{O_{2}} = 32, R = \frac{25}{3} J - m o l^{- 1} K^{- 1})$ .

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$385 N m^{- 2}$

$38.5 N m^{- 2}$

$3850 N m^{- 2}$

$3.85 N m^{- 2}$

answer is C.

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By Dalton’s law,

$\begin{array}{l} P = P_{H_{2}} + P_{O_{2}} \\ = \frac{n_{H_{2}} R T}{V} + \frac{n_{O_{2}} R T}{V} = (\frac{0.5}{2} + \frac{0.8}{32}) (\frac{25}{3}) (\frac{300}{(\frac{5}{28})}) = 3850 {Nm}^{- 2} \end{array}$

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A vessel contains 0.5 g of hydrogen and 0.8 g of oxygen. The volume of vessel is 528 m3 while its temperature is maintained at 300 K. Then the pressure of the mixture is MH2=2, MO2=32, R=253J−mol−1K−1.

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A vessel contains 0.5 g of hydrogen and 0.8 g of oxygen. The volume of vessel is $\frac{5}{28} m^{3}$ while its temperature is maintained at 300 K. Then the pressure of the mixture is $(M_{H_{2}} = 2, M_{O_{2}} = 32, R = \frac{25}{3} J - m o l^{- 1} K^{- 1})$ .