A quantity of hydrogen gas occupies a volume of `30.0 mL` at a certain temperature and pressure. What volume would half of this mass of hydrogen occupy at triple the initial temperature, if the pressure was one-ninth that of the original gas?

Options

A. 270 mL
B. 90 mL
C. 405 mL
D. 135 mL

Answer

C

Explanation

`PV = nRT`
`P xx 30 = nRT`....(i)
In second case mass of hydrogen is `1//2` of its original mass, i.e., the gas contains `n//2` moles of `H_(2)`. In second case, pressure is `P//9` and temperature is 3 T. Let volume in second case is V'
`:. (P//9) xx V' = (n//2) xx R xx (3T)`
`(PV')/(9) = (3nRT)/(2)` ....(ii)
Dividing (ii) by (i)
`(V')/(9 xx 30) = (3)/(2)`
`V' = (30 xx 9 xx 3)/(2) = 405 mL`