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A cylinder containe 20 kg of `N_(2)` gas (M= 28 kg `K^(-1) mol^(-1))` at a [ressire pf 5 atm. The mass of hydrogen (M = 28 kg `K^(-1) mol^(-1))` at a pressure of 3 atm contained in the same cylinder at same temperature isA. 1.08 kgB. 0.86 kgC. 0.68 kgD. 1.68 kg |
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Answer» Correct Answer - B `V/T=(nR)/p` V and T for both cases in same. Hence `(n_(1))/(p_(1))=(n_(2))/(p_(2))or (m_(1))/(p_(1)M_(1))=(m_(2))/(p_(2)M_(2))` or `" "m_(2)(p_(2)M_(2))/(p_(1)M_(1)).m_(1)=((3)(2))/((5)(28)).20=0.86kg` |
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