Question

Two identical pistons begin and end in the same state. Initially, they contain 0.04-moles of gas, are at room temperature (20°C), and have a volume of 0.03- m3.

The first is compressed isothermally until its volume is halved. The second goes through a two-step process: First the pressure is changed isochorically, then the volume is changed isobarically.

A) draw a P-V diagram for each of these processes

B) Find the final state (P, V, and T) of the two systems.

C) Find and compare the work done in each process.

Answer 1

(A) The PV diagram of the given processes are shown in the following figure

The initial state of the gases 1 and 2 is A. The first gas goes through path A to C (shown as green curve), while the second gas goes through path A to B to C (shown in red). A to C is an isothermal process, the temperature along the curve is . A to B is isochoric process, the volume of the gas is in this process. B to C is an isobaric process, the pressure is fixed at (which will be identified in the next part).

(B) Given that the volume is changed to half of its initial value in the final state. Therefore,

States A and C lies along the isothermal curve of temperature . Therefore,

Using ideal gas equation we have

Substituting we get

The final state has

(C)

The work done by the first gas is equal to the work done in an isothermal process. We know that work done in an isothermal expansion of a gas from volume to at temperature is

The work done by the first gas is

Substituting and we get

The work done by secod gas is equal to the sum of the work done in isobaric and isochoric process.

The work done in an isochoric process is zero because there is no change in the volume of the gas. The work done in changing volume of a gas from at pressure P is .Therefore

The negative work done means that the work is done by us on the gas. The work done by us on the first gas is smaller than the work done on gas 2.