The gas in an air cannon expands as shown below, from V1 = 2.0-m3 to V2 = 8.0-m3, while the pressure changes from P1 = 6.0-atm to P2 = 2.0-atm. If the heat supplied to the gas is 1.2 × 106-J and 1-atm = 1.013 × 105-N/m2, what is the change in the internal energy of the gas (in J, and watch out for signs) b). Calculate the change in energy if the above process occurs in two steps. The first step is an isovolumetric change from P1 to P2, and the second an isobaric expansion from V1 to V2.

Answer:

In this expansion, pressure changes from 6.0 atm to 2.0 atm, while volume change from 2.0 m³ to 8.0m³

It is an irreversible expansion, as it occurs rapidly.

Work done in an irreversible expansion process W_irr = - P2(V2 - V1) = - 2 atm (8-2) m³

= -12 m³ atm = -12x10³ L atm

1 L atm = 101.3 J

Therefore, -12x10³ L atm = 101.3J x -12x10³ L atm / 1 L atm = -1.215x10⁶ J work is done by air

Heat supplied = 1.2x 10⁶J

Change in internal energy = q + w = 1.2x10⁶  -1.215x10⁶ = 0

In other words, the process is isothermal, that is temperature remains constant during the process

The gas spends its energy during expansion, and therefore it temperature decreases. in order to keep T constant energy in the form of heat is absorbed from the surroundings

If the process occurs at constant volume, that is isovolumetric, no work is done. therefore, change in internal energy = heat supplied = 1.2x10⁶J

In an isobaric process, pressure remains constant and there will be a change in volume

Thus pressure remains at 2 atm that is external pressure

w = - p_ext(V₂-V₁) = -2 atm (8-2) m³ = -1.215x10⁶J

Change in energy = change in energy in the first step + in the second step = 0

This indicates that change in internal energy is a state function. That is, depends upon the initial and final state and is indpendent of the number of steps taken to reach there.