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 is

Options

A. 1.08 kg
B. 0.86 kg
C. 0.68 kg
D. 1.68 kg

Answer

B

Explanation

`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`