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 "joules"`
B. `300 R "joules"`
C. `75 R "joules"`
D. `100 R joules`

Answer

B

Explanation

For one mole of gas
`Delta Q = C_v DeltaT + P DeltaT`
At constant volume `DeltaT = 0`
for two moes of gas,
`DeltaT = 0`
From `PV = nRT = 2R xx 100`
and `P/2 V = 2RT_(f)`
`:. T_(f) = 150K`
`:. DeltaQ = 2C_(v)(T_f-T_i) = 2C_v(150-300)`
`=-300 C_v` joules
In the next process,
`Delta = 2C_p DeltaT=2C_p(300-150)`
`=300C_p` joules.
`:.` Net heat absorbed = `-300C_v+300C_p`
`=300(C_p-C_v)`
`=300R "joules"`.