Assume ideal gas behaviour for all the gases considered and neglect vibrational degrees of freedom. Separate equimolar sample. 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 pressue.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 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 sample. 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 pressue.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 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 minimum for `SO_(2)` There will be no change in INTERNAL energy for any of the GAS after both the steps of process are competed 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_("toal")=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))lt gamma_(CO_(2))=gamma_(O_(2))=ltgamma_(Ne)` so, max. decrease in temp. of Ne due to step 1.