InterviewSolution
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InterviewSolution
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A vessel filled with gas is divided into two equal parts `1` and `2` by a thin heat-insulating partition with two holes. One hole has a small diameter, and the other has a very large diameter (in comparsion with the mean free path of molecules). In part `2` the gas is kept at a temperature `eta` times higher than that of part `1` How will the concentration of molecules in part `2` change and how many times after the large hole is closed ? |
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Answer» If the temperature of the compartment `2` is `eta` times more than that of compartment `1`, it must contain `(1)/(eta)` times less number of molecules since pressure must be the same when the big hole is open. If `M` = mass of the gas in `1` than the mass in `1` than the mass of the gas in `2` must be `(M)/(eta)`. So immediately after the big hole is closed. `n_1^0 = (M)/(mV), n_2^0 = (M)/(m V eta)` where `m` = mass of ach molecule and `n_1^0, n_2^0` are concentrations in `1` and `2`. After the big hole is closed the pressures will differ and concentration will become `n_1` and `n_2` where `n_1 + n_2 = (M)/(m V eta) (1 + eta)` On ther other hand `n_1 lt v_1 gt = n_2 lt v_2 gt` i.e. `n_1 = sqrt(eta) n_2` Thus `n_2 (1 + sqrt(eta)) = (m)/(m V eta) (1 + eta) = n_2^0 (1 + eta)` So `n_2 = n_2^0 (1 + eta)/(1 + sqrt(eta))`. |
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