# Isothermal thermodynamic processes – problems and solutions

1. PV diagram below shows an ideal gas

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