Work done in thermodynamics process – problems and solutions

1. An ideal gas is compressed from state 1 to state 2 as shown in the graph below. Determine the work done by the gas.

Solution

1 liter = 0.001 m^{3 }

Known :

Pressure 1 (P_{1}) = 10 kPa = 10,000 Pa = 10,000 N/m^{2}

Pressure 2 (P_{2}) = 20 kPa = 20,000 Pa = 20,000 N/m^{2}

Volume 1 (V_{1}) = 4 liters = 4 (0.001 m^{3}) = 0.004 m^{3}

Volume 2 (V_{2}) = 10 liters = 10 (0.001 m^{3}) = 0.010 m^{3}

__Wanted :__ The work done by the gas (W)

__Solution :__

W = P (V_{2} – V_{1}) + 1/2 (P_{2} – P_{1})(V_{2} – V_{1})

W = 10,000 (0.010-0.004) + 1/2 (20,000-10,000)(0.010-0.004)

W = 10,000 (0.006) + 1/2 (10,000)(0.006)

W = 10 (6) + 1/2 (10)(6)

W = 60 + 1/2 (60)

W = 60 + 30

W = 90 Joule

2. Determine the work done by the gas in process a-b-c-d-a, as shown in graph below.

Known :

Pressure 1 (P_{1}) = 1 x 10^{5} Pa = 1 x 10^{5} N/m^{2}

Pressure 2 (P_{2}) = 3 x 10^{5} Pa = 3 x 10^{5} N/m^{2}

Volume 1 (V_{1}) = 2 m^{3}

Volume 2 (V_{2}) = 4 m^{3}

__Wanted :__ Work (W)

__Solution :__

The work done by the gas = area of rectangular a-b-c-d.

W = (P_{2} – P_{1})(V_{2} – V_{1})

W = (3 x 10^{5} – 1 x 10^{5})(4 – 2)

W = (2 x 10^{5})(2)

W = 4 x 10^{5} Joule

W = 400,000 Joule

W = 400 kJ

3. Determine the work done by the gas in process A-B-C-A.

__Known :__

Pressure 1 (P_{1}) = 1 N/m^{2}

Pressure 2 (P_{2}) = 7 N/m^{2}

Volume 1 (V_{1}) = 1 m^{3}

Volume 2 (V_{2}) = 5 m^{3}

__Wanted :__ Work (W)

__Solution :__

The work done by the gas = area of triangle A-B-C

W = ½ (P_{2} – P_{1})(V_{2} – V_{1})

W = ½ (7-1)(5-1)

W = ½ (6)(4)

W = ½ (24)

W = 12 Joule

**What is the definition of work in thermodynamics?****Answer:**In thermodynamics, work is defined as an energy transfer to or from a system that is not due to a temperature difference between the system and its surroundings.

**How is work related to the pressure-volume (P-V) diagram of a thermodynamic process?****Answer:**In a P-V diagram, the work done by or on a system during a process is represented by the area under the curve of the process. For processes involving gases, the work is often given by the integral of P dV, where P is the pressure and dV is the differential change in volume.

**What is the work done during an isochoric process?****Answer:**An isochoric process occurs at constant volume. Since there’s no volume change, the work done during an isochoric process is zero.

**How does the work done in an isobaric (constant pressure) expansion differ from that in an isothermal (constant temperature) expansion of an ideal gas?****Answer:**In an isobaric expansion, work is calculated as $W=P×ΔV$, where P is the constant pressure and $ΔV$ is the change in volume. For an isothermal expansion of an ideal gas, work can be calculated using the equation $W=nRTln(V )$, where n is the number of moles, R is the universal gas constant, and $V_{f}$ and $V_{i}$ are the final and initial volumes, respectively.

**What is the significance of the first law of thermodynamics regarding work and energy?****Answer:**The first law of thermodynamics states that the change in the internal energy of a system is equal to the heat added to the system minus the work done by the system on its surroundings. It represents the conservation of energy principle.

**For a cyclic process on a P-V diagram, what does the enclosed area represent?****Answer:**For a cyclic process on a P-V diagram, the enclosed area represents the net work done by or on the system during one cycle.

**How does the work done in an adiabatic process differ from an isothermal process for an ideal gas?****Answer:**In an adiabatic process, there is no heat exchange with the surroundings, so the work done results in a change in the internal energy of the gas. For an isothermal process, the temperature remains constant, and any work done is balanced by an equal amount of heat transfer, so there’s no change in internal energy.

**Why can the work done in an irreversible process be different from a reversible process even if the initial and final states are the same?****Answer:**In a reversible process, each intermediate state between the initial and final states is in equilibrium. In an irreversible process, this isn’t the case. The paths taken on a P-V diagram, and thus the work done, can differ significantly between the two types of processes, even if they start and end at the same points.

**How does the presence of non-ideal (real) gas behavior affect work calculations in thermodynamic processes?****Answer:**For real gases, deviations from ideal behavior can lead to differences in work calculations, especially at high pressures or low temperatures. The Van der Waals equation or other real gas models might need to be used to account for these deviations.

**What does it mean when the work done on a system is negative?**

**Answer:**When the work done on a system is negative, it means the system is doing work on its surroundings. This often corresponds to an expansion of a gas against external pressure.