Answer to Question #139766 in Molecular Physics | Thermodynamics for Zep Santos

Question

 A gas in a piston-cylinder assembly undergoes two processes in series. From state I to state 2 then: is  energy transfer by heat to the gas of 500 kJ and the gas does work on the piston amounting 800 kJ. The second process. from state 2 to state 3 is a constant pressure compression at 400 kPa, during which  there is a heat transfer from the gas amounting 450 kJ. The following data are also known: U1 = 2000 kJ  and U3 = 3500 kJ. Neglecting changes in KE and PE, calculate the change in volume of the gas during  process 2-3.

Answer 1

For the state 1 to state 2 :

U1 = 2000 kJ

Heat added to gas, q = + 500 kJ

Work done by the gas, w = – 800 kJ

(Conventionally work done by the gas is negative)

Let at state 2 Internal energy = U2 kJ

∆U = q + w

Or, (U2–U1) = q + w

Or, U2 = q + w + U1

Putting the values we get

Or, U2 = (500 – 800 + 2000) kJ

= 1700 kJ

For state 2 to state 3 :

U2 = 1700 kJ

U3 = 3200 kJ

heat transfer from the gas,q = – 450 kJ

Pressure,P = 400 kPa

Let, work done = w

= –P∆V

Where, ∆V = change in volume

Similarly,

∆U = q + w

Or, (U3–U2) = q + w

Or, w = U3 – U2 –q

Or, –P∆V = 3200 – 1700 +450

Or, –P∆V = 1950

Or, ∆V = –(1950 kJ/400 kPa)

= – 4.875 m³

Hence,

change in volume of the gas during process(2-3), ∆V = – 4.875 m³