Assume ideal gas behaviour for all the gases considered and neglect vibrational degrees of freedom. Separate equimolar samples of Ne, O_(2), CO_(2) and SO_(2) were subjected to a two process as mentioned. Initially all are at same state of temperature and pressure Step I rarr All undergo reversible adiabatic expansion to attain same final volume, which is double the original volume thereby causing the decreases in their temperature Step II rarr After step I all are given appropriate amount of heat isochorically to restore the original temperature Mark the correct option(s)

Assume ideal gas behaviour for all the gases considered and neglect vi

1.	Assume ideal gas behaviour for all the gases considered and neglect vibrational degrees of freedom. Separate equimolar samples of Ne, O_(2), CO_(2) and SO_(2) were subjected to a two process as mentioned. Initially all are at same state of temperature and pressure Step I rarr All undergo reversible adiabatic expansion to attain same final volume, which is double the original volume thereby causing the decreases in their temperature Step II rarr After step I all are given appropriate amount of heat isochorically to restore the original temperature Mark the correct option(s)
Answer» Due to step I only, the decrease in temperature will be maximum for Ne During step II, heat given will be maximum for `SO_(2)` There will be no change in internal energy for any of the gas after both the steps of process are completed The P-V graph of `O_(2)` and `CO_(2)` will be same Solution :First step is adiabatic (q=0) so `DeltaU_(1)=w_(1)` SECOND step is isochoric (w=0) So, `DeltaU_(2)=q_(2)` `because` initial and FINAL temp. are same `therefore DeltaU_("TOTAL")=DeltaU_(1)+DeltaU_(2)=0 or w_(1)+q_(2)=0` Max, work done by the gas, `SO_(2)` is (area) under the curve so, `SO_(2)` absorbed `because gamma_(SO_(2))ltgamma_(CO_(2))=gamma_(O_(2))ltgamma_(Ne)` so, max. decrease in temp. of Ne due to step 1.