Question

One mole of an ideal gas at 1.0 Mpa and 300K is heated at constant pressure till the volume is doubled and then it is allowed to expand at constant temperature till the volume is doubled again. Calculate the work done by the gas.

Answer 1

Amount of Gas = 1 mole

P1 = 1.0 MPa

T1 = 3000K Process 1–2: Constant pressure

P1V1/T1 = P2V2/T2 i.e.; V1/T1 = V2/T2

V2 = 2 V1; i.e.; V1/300 = 2V1/T2

T2 = 600K ...(i)

For 1 mole, R = Universal gas constant

= 8.3143 KJ/kg mole K

= 8314.3 Kg–k

= PV2 – PV1

= R(T2 – T1) = 8314.3 (600 – 300) = 2494.29KJ ...(i)

Process 2 – 3: Isothermal process

= P2V2ln V3/V2 = RT2ln 2V2/V2 = RT ln 2 = 8314.3 × 600 ln 2 = 3457.82KJ

Total WD = WD1–2 + WD2–3

= 2494.29 + 3457.82 = 5952.11 KJ

=5952.11J