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`4 g` hydrogen is mixed with 11.2 litre of He at (STP) in a container of volume 20 litre. If the final temperature is `300 K`, find the pressure. |
Answer» `4g` hydrogen `=2` moles hydrogen `11.2` litre He at `STP = (1)/(2)` mole of He `P = P_(H) + P_(He) = (n_(H) + n_(He))(RT)/(V)` `= (2 + (1)/(2))(8.31 xx (300))/((20 xx 10^(-3))) = 3.12 xx 10^(5) N // m^(2)` |
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