# Isothermal thermodynamic processes – problems and solutions

1. PV diagram below shows an ideal gas undergoes an isothermal process. Calculate the work is done by the gas in the process AB.

Solution

Work done by a gas is equal to the area under the PV curve

AB = triangle area + rectangle area

W = [½ (8 x 10^{5}–4 x 10^{5})(3-1)] + [4 x 10^{5} (3-1)]

W = [½ (4 x 10^{5})(2)] + [4 x 10^{5} (2)]

W = [4 x 10^{5}] + [8 x 10^{5}]

W = 12 x 10^{5} Joule

The work is done by the gas in the process AB = 12 x 10^{5} Joule

2. Calculate the work is done by an ideal gas in the process ABC.

The work is done by an ideal gas in the process ABC = the area under the PV curve

AB = triangle area + rectangle area

W = [½(10×10^{5}–5×10^{5})(30-10)]+[5×10^{5}(30-10)]

W = [½ (5 x 10^{5})(20)] + [5 x 10^{5} (20)]

W = [(5 x 10^{5})(10)] + [100 x 10^{5}]

W = [50 x 10^{5}] + [100 x 10^{5}]

W = 150 x 10^{5} Joule

W = 1.5 x 10^{7} Joule

3. An ideal gas undergoing isothermal processes. What is an amount of heat is added to the gas so the gas do work of 5000 Joule on the environment.

__Known :__

Work (W) = 5000 Joule

__Wanted:__ Heat is added to the gas (Q)

__Solution :__

An isothermal process is a thermodynamic process that occurs at a constant temperature.

ΔU = 3/2 n R ΔT

*ΔU **= the change in internal energy, n = number of moles, R = universal gas constant, **ΔT = The change in temperature.*

According to the above equation, if ΔT = 0 then ΔU = 0.

__The equation of the first law of thermodynamics :__

ΔU = Q – W

0 = Q – W

Q = W

Q = 5000 Joule.

4. PV diagram for an ideal gas undergoing isothermal process shown in the figure below. Calculate the heat is added by a gas in process AB.

__Known :__

Pressure 1 (P_{1}) = 5 atm = 5 x 10^{5} Pa

Pressure 2 (P_{2}) = 10 atm = 10 x 10^{5} Pa

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

Volume 2 (V_{2}) = 6 m^{3}

__Wanted__ : Heat is added in process AB.

__Solution :__

Isothermal = constant temperature. According to the equation below, if ΔT = 0 then ΔU = 0.

ΔU = 3/2 n R ΔT

ΔU = 3/2 n R (0)

ΔU = 0

Apply to the first law of thermodynamics :

ΔU = Q-W

0 = Q-W

Q=W

The work is done by an ideal gas = the area under the PV curve = triangle area + rectangle area

W = ½ (P_{2} – P_{1})(V_{2} – V_{1}) + P_{1 }(V_{2} – V_{1})

W = ½ (10 x 10^{5} – 5 x 10^{5})(6-2) + (5 x 10^{5})(6-2)

W = ½ (5 x 10^{5})(4) + (5 x 10^{5})(4)

W = ½ (20 x 10^{5}) + (20 x 10^{5})

W = (10 x 10^{5}) + (20 x 10^{5})

W = 30 x 10^{5 }Joule