Question

At `27^@C` two moles of an ideal monoatomic gas occupy a volume V. The gas expands adiabatically to a volume 2V. Calculate (i) the final temperature of the gas, (ii) change in its internal energy, and (iii) the work done by the gas during this process.

Answer 1

(i) Here, `T_(1)= 27^(@)C= (27+273)K= 300K`
`n=2, T_(2)=? V_(2)=2V_(1)`
In an adiabatic change
`T_(2)V_(2)^(gamma-1)= T_(1)V_(1)^(gamma-1)`
`T_(2)=T_(1)((V_(1))/(V_(2)))^(gamma-1)= 300(1/2)^(5//3-1)`
`T_(2)= 300(0.5)^(2//3)= 189K`
(ii) Change in internal energy
`dU=nC_(v).dT`
As `C_(v)=(R)/(gamma-1)`
`:. dU=(nR.dT)/(gamma-1)`
`=(2xx8.31(189-300))/((5//3-1))`
`= -2xx8.31xx111xx3/2`
`= - 2767.23J`
(iii) According to first law of thermodynamics,
`dQ=dU+dW`
In an adiabatic process, `dQ=0`
`:. dW= -dU= -(-2767.23)J`
`2767.23J`