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

Options

A. `150 R J`
B. `300 R J`
C. `75 R J`
D. `100 R J`

Answer

B

Explanation

b. For 1 mol of gas,
`Delta Q = C_(v) Delta T + P Delta T`
At constant volume, `Delta T = 0`
For 2 moles of gas,
`Delta = 2 C_(v) Delta T`
From `PV = nRT = 2R xx 300`
and `(P)/(2) V = 2 RT_(f)`
`:. T_(f) = 150 K`
`:. Delta Q = 2 C_(V) (T_(f) - T_(i)) = 2 C_(V) (150 - 300)`
`= - 300 C_(v) J`
In the next process,
`Delta Q = 2 C_(P) Delta T = 2 C_(P) (300 - 150)`
`= 300 C_(v) J`
`:.` Net heat absorbed `= - 300 C_(v) + 300 C_(P)`
`= 300 (C_(P) - C_(v)) = 300 R J`