A massless piston divides a closed, thermally insulated cylinder into two equal parts. One part contains $$m=28$$ gram of nitrogen at temperature $$T0$$. At this temperature $$one-$$ third of molecules are dissociated into atoms and the other part is evacuated. The piston is released and the gas fills the whole volume of the cylinder. Then the piston is slowly displaced back to its initial position. The increase in internal energy of the gas is: (Neglect$$ further dissociation of molecules during motion of the $$piston)

Question

A massless piston divides a closed, thermally insulated cylinder into two equal parts. One part contains $$m=28$$ gram of nitrogen at temperature $$T0$$. At this temperature $$one-$$ third of molecules are dissociated into atoms and the other part is evacuated. The piston is released and the gas fills the whole volume of the cylinder. Then the piston is slowly displaced back to its initial position. The increase in internal energy of the gas is: (Neglect$$ further dissociation of molecules during motion of the $$piston)

Infinity Learn · Accepted Answer

Molar mass of $$N2=$$ $$M=28$$ g $$mol-1$$  Number of moles of $$N2$$,$$n0=mM=2828=1$$ mol Since,  $$One-third$$ molecules are dissociated into atoms.   $$N2$$ →$$2N$$ ∴ No. of moles of diatomic $$N2=23mol$$ and  No. of moles of monatomic nitrogen $$=2$$×$$13=23$$ mol CV $$mix=n1Cv1+n2Cv2n1+n2=2332R+2352R23+23=2R$$ ∴CP $$mix=CV$$ $$mix+R=2R+R=3R$$ ∴γ$$mix=CP$$ mixCV $$mix=1.5$$ Now,  Piston is diaplaced back to its initial position, and during the adiabatic  compression, volume of the gas mixture is halved.  ∴$$T1V1$$γ$$-1=T2V2$$γ$$-1$$ ⇒$$T0V0$$γ$$-1=T2V02$$γ$$-1$$ ⇒$$T2=T02$$ Change in Internal $$energy=$$ Δ$$U=n1+n2CVT2-T0=23+23$$×$$2RT02-T0=83RT0(2-1)$$

Remember concepts with our Masterclasses.

Ready to Test Your Skills?

free study material