Two moles of helium gas`(lambda=5//3)` are initially at temperature `27^@C` and occupy a volume of 20 litres. The gas is first expanded at constant pressure until the volume is doubled. Then it undergoes an adiabatic change until the temperature returns to its initial value.
(i) Sketch the process on a p-V diagram.
(ii) What are the final volume and pressure of the gas?
(iii) What is the work done by the gas ?

Answer

`0.44xx10^5Pa;12.45 K`

Explanation

`n_(He)`=2"mole"`(r=5//3)`
`T_(1)=27^(@)C=300K`,`V_(1)=20 liters`, P
`downarrow`" isobaric process"`(v)/(T)=constant`
P,v=40 liter, `T_(2)=600 K`
`downarrow` "adiabatic"`(triangleQ)=0`
`P_(3),V_(3),T_(3)=300K` (b) inital pressure `P_(1)=(nRT)/(v)=(2xx25xx300)/(3xx20xx10)`
`1liter = 10^(-3)=10^(-3)m^(3)`
`P_(1)=250xx10^(3)`
`=2.5 atm`
A(2.5 atm,20 it, 300K) `downarrow`
B(2.5)atm,40 It,600 k
`downarrowPV^(r)=constant`
`C(P_(3),V_(3)300)`
`PV^(r)=const`
`TVR^(r-1)=constant`
`T_(2)V_(2)^(r-1)=T_(3)V_(3)^(r-1)`
`600xx(40)^(r-1)=(300)(V_(3))^(r-1)`
`2(40)^(2//3)=(V_(3)).40=80sqrt(2)ltr`
`PV^(r)=constatn`
`P_(2)V_(2)^(r) =P_(3)V_(3)^(r)`
`(2.5)(40)^(5//3)= P_(3)(80sqrt(2))^(5//3)`
`P_(3) = 2.5 xx((1)/(2sqrt(2)))^(5//3)`
`=0.44 atm`
(c) `W_(ArarrB) +W_(BrarrC) = total work done`
`W_(A rarr B) = P triangle V = 2.5 xx 10^(5) xx(20 xx10^(-3)) =5xx10^(3) = 5ks`
`W_(B rarr C) = (P_(i)V_(i) - P_(f)V_(f))/(r-1)`
`(P_(2)V_(2) - P_(3)V_(3))/(r-1)`
`(2.5xx40-0.44xx80sqrt(2)xx10^(2))/(2//3)`
`(3)/(2)(100-49.78)xx10^(2) = 7.5 kJ`
`W_"(net)" = 5+7.5 =12.5 kJ`