Two moles of helium gas undergo a cyclic process as shown in Fig. Assuming the gas to be ideal, calculate the following quantities in this process

(a) The net change in the heat energy
(b) The net work done
(c) The net change in internal energy

Answer

A, B, C

Explanation

Let us find out the work done in the cycle

Work done from A to B (Isobaric process)
`W_(AB)=nR(T_B-T_A)`
`=nRxx100=2xx200xx8.32=1664J`
Work done from B to C (Isothermal process)
`W_(BC)=2.303nRTlog_(10)(P_B)/(P_C)`
`=2.303nRxx400log_(10)2/1=277.2nR`
`=554.4xx8.32=4612.6`
Work done from C to D (Isobaric process)
`W_(CD)=nR(T_D-T_C)=nR(300-400)`
`=-100nR=-200xx8.32=-1664J`
Work done from D to A (Isothermal process)
`W_(DA)=2.303nRTlog_(10)(P_D)/(P_A)=2.303nRxx300log_(10)1/2`
`=-207.9nR`
`=-415.8xx8.32=-3459.5J`
`The total work done =W_(AB)W_(BC)W_(CD)W_(DA)`
` =1153J`
(a) `DeltaU=Q-W`
For compleat cycle `DeltaU=0`
`:. Q=W=1153J`
(b) `W=1153J`
(c) `DeltaU=0.` Since, the process is cyclic.