1 mol of an ideal gas is expanded from volume `V_(1) " to " 10 V_(1)` . If work done by the gas is 10 kJ and the initial pressure of the gas is `1xx10^(7) Pa`, then calculate `V_(1)`. What will be the temperature of the gas if the amount is 2 mol ?

Explanation

`w=- nRT_(1)xx2.303 " log " (V_(2))/(V_(1))`
`= - 1xx8.314xxT_(1)xx2.303 " log " (10V_(1))/(V_(1))`
or, `10xx1000= - 8.314xxT_(1)xx2.303 " or " T_(1)=522.27 K`
Again, `P_(1)V_(1)= nRT_(1)=1 xx RT_(1)`
or , `V_(1)=(RT_(1))/(P_(1))= (0.0821 " L atm" xx 522.27)/(10^(7) P_(a))`
`= (0.0821xx522.2xx1.013xx10^(5))/(10^(7))L=0.434 L`
And , `P_(1)V_(1)= nRT_(2)= 2xx RT_(2) therefore T_(2)=260.9 K`