Two moles of an ideal gas at temperature `T_(0) = 300 K` was cooled isochorically so that the pressure was reduced to half. Then, in an isobaric process, the gas expanded till its temperature got back to the initial value. Find the total amount of heat absorbed by the gas in the processs

Explanation

In an isochoric process, `Delta W = 0` `(Delta V = 0)`
and `Delta U = C_(V) Delta T = (R )/(gamma - 1) Delta T` `( :' C_(v) = (R )/(gamma - 1))`
`implies Delta U = (Delta (RT))/(gamma - 1) = (Delta (pV))/(gamma - 1)`
`= (V)/(gamma - 1) Delta P = (V)/(gamma - 1) ((1)/(2) P - P) = - (1)/(1) (p V)/(2 gamma - 1)`
Now `Delta Q = Delta U + Delta W`
`:. Delta Q = - (1)/(2) (p V)/(gamma -1) + 0 = - (1)/(2) (p V)/(2 gamma - 1) = (1)/(2 gamma) (RT)/(-1)`
The negative sign shows that heat is not added but subtracted form the gas in the process
In the isobaric process
`Delta W = int_(V_(i))^(V_(f)) dV = (P)/(2) (V_(f) - V_(i))` [ In this process pressure is `(P)/(2)`]
In an isochroic process, V remain constant, so `p prop T` and temperature is reduced to `T_(0)//2`
In the isobaric process, the original temperature is restored.
`Delta T = T_(0) - T_(0)//2 = T_(0)//2`
`Delta U C_(v) Delta T = (R )/(gamma - 1) (T_(0))/(2) ( :' C_(V) = (R )/((gamma - 1))` and `Delta T = (T_(0))/(2))`
`:. Delta Q = Delta U + Delta W = (1)/(2) (RT)/(gamma -1) + (1)/(2) RT_(0)`
`:.` Net heat added is
`[-(1)/(2) (RT_(0))/(gamma - 1)] + [(1)/(2) (RT_(0))/(gamma - 1) + (1)/(2) RT_(0) ] = (1)/(2) RT_(0)`
This is the heat added when the number of moles is one.
When there are 2 moles, heat added is
`2 xx (22)/(7) RT_(0) = 8.3 xx 300 = 2490 J`