An ideal gas is compressed to half of the volume. How much work is done if the process of compression is (a) Isothermal (b) Adiabatic ? In which of the two cases will the work done by more ? How do you account for the difference ?

An ideal gas is compressed to half of the volume. How much work is don

1.	An ideal gas is compressed to half of the volume. How much work is done if the process of compression is (a) Isothermal (b) Adiabatic ? In which of the two cases will the work done by more ? How do you account for the difference ?
Answer» Solution :(a) In case of isothermal change `W=nRT log_(e) (V_(F)//V_(I))` So `W_(1)=nRTlog_(e) [1//2] = -0.6930nRT "…….." (1)` (b) In case of ADIABATIC change, `W=(nR[T_(F)-T_(I)])/(1-gamma)" with "TV^(gamma-1)=constant` i.e.,`W=(nRT_(1))/(1-gamma)[1-T_(F)/T_(I)]" with " T_(F)/T_(I)=[V_(I)/V_(F)]^(gamma-1)` so `W=(nRT)/([(5//3)-1])[1-2^(2//3)] ["as "gamma =5/3,T_(1)=T]` or `W_(A) = -3/2 nRTxx0.5874=-0.88 nRT "......."(2)` Negative sign means that work is done on the gas. COMPARING Eqns. (1) and (2) we find that for same COMPRESSION `W_(A) gt W` , i.e., more work is required in adiabatic compression than in isothermal. In adiabatic compression, temperature and hence internal energy of THEGAS also increase and so more work will be required in comparison to isothermal compression in which temperature and hence internal energy remains constant.