Question

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 1

As we know in a process, the change in head ererngy of heat shpplied to the gas is equal to the net work done by the gas
Here, AB is isobaric process. Hence, work done during this process form A to B is
`W_(AB) = P (V_(2) - V_(1)) = nRT (T_(2) - T_(1))`
or `W_(AB) = 2 xx 8.314 xx (400 - 300) = 1662.8 J`
Work done during isothermal process form B to C is
`W_(BC) = nRT_(C ) 1n (V_(2) // V_(1)) = nRT+(C ) 1n (P_(1) // P_(2))`
`= 2xx 8.314 xx 400 xx 1n (2) = 2 xx 8.314 xx 400 xx 0.693`
`= 4601.2 J`
Work done during isobaric process from C to D
`W_(CD) = nR (T_(D) - T_(C )) = 2 xx 8.314 xx (300 - 400)`
`= - 1662.8 J`
Work done during isothermal process from D to A
`W_(DA) = nRT_(D) 1m (P_(D)//P_(A))`
`= nRT_(D ) 1n (2)`
`= - 2 xx 8.314 xx 300 x 0.693`
`= - 3437.7 J`
Net down done
`= W_(AB) + W_(BC) + W_(CD)+ W_(DA)`
` = 1662.8 + 4610.2n- 1662.5 = 3457.7 = 1152.5 J`
Now from the first law of thermodynamics
`Delta Q = DeltaU + Delta W`
Here, `Delta Q = Delta W = 1152.5 J`
`So the heat given to the system is 1152.5 J
As the gas returns to its original state, there is no change in internal energy.