Pressure exerted by 12g of an ideal gas at `t^(@)C` in a vessel of volume VL is 1 atm. When the temperature is increased by `10^(@)C` at the same volume, the pressure increased by `10%`. Calculate the temperature 't' and volume V (molar mass of gas=120).

Explanation

12g of the gas=`(12)/(120)=0.1` mol of the gas
Given that the initial pressure of the gas, P=1 atm and its initial temperature, `T=(273+t)K`
by applying ideal gas equation, PV=nRT, we get
`1xxV=1xx0.0821xx(273+t)` . . [1]
It is given, final pressure of the gas, `P=(1+(10)/(100)xx1)=1.1` atm
and its final temperature `=(273+t+10)K(283+t)K`
By applying ideal gas equation, PV=nRT, we get
`1.1xxV=1xx0.0821x(283+t)` . . [2]
from the equations [1] and [2], we have
`1.1xxV=1xx0.0821xx(283+t)`
From the equations [1] and [2], we have
`1.1xx0.0821xx(273+t)=1xx0.0821xx(283+t)`
or, `1.1(273+t)=283+t`
`therefore t=-173,` so, the value of `t=-173^(@)C`
Puttingt=-173 in equation [1] (or equation [2]),
We have, `V=1xx0.0821(273-173)=0.821L`.