Two moles of an ideal gas at `300 K` cooled at constant volume so that the pressure is reduced to half the initial value. Then as a result of heating at constant pressure. Find the total heat absorbed by the gas, if `R` is the gas constant.

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`