The pressure exerted by 12 g of an ideal gas at temperature `T(""^(@)C)` in a vessel of volume V is 1 atm. When the temperature is increased by `10^(@)C` at the same volume, the pressure increases by 10%. If molecular weight of the gas is 120, the temperature `(T^(@)C)` and volume (V) are

Options

A. `T= -273^(@)C, V= 0.082 L`
B. `T= -173^(@)C, V= 0.82 L`
C. `T= 0^(@)C, V= 22.4 L`
D. `T= 27^(@)C, V= 22.4L`

Answer

B

Explanation

Moles of gas `(12)/(120)= 0.1`
`pV= nRT" or "1 xx V= 0.1 xx R(273+t)" ""…………"(1)`
After temperature change, `p= 1+(10)/(100)= 1.1 atm`
`T= (t+273+10)= (283+t)K`
`1.1 xx V= 0.1 xx R xx (283+t)" ""………….."(2)`
Dividing Eq. (2) by Eq. (1), `(1.1)/(1)= (283+t)/(273+t) implies t= -173^(@)C`
substituting the value of t in Eq. (2), `1.1 xx V= 1(283-173)xx 0.082 implies V= 0.82L`.