# Isobaric thermodynamics processes – problems and solutions

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

__Known :__

Pressure (P) = 5 x 10^{5} N/m^{2}

Initial volume (V_{1}) = 2 m^{3}

Final volume (V_{2}) = 6 m^{3}

__Wanted :__ Work (W)

__Solution :__

W = P (V_{2} – V_{1})

W = (5 x 10^{5})(6 – 2) = (5 x 10^{5}) (4)

W = 20 x 10^{5 }= 2 x 10^{6} Joule

2. What is difference of the work is done by the gas in process AB and process CD…

__Known :__

**Isobaric process ****AB** :

Pressure (P) = 6 atm = 6 x 10^{5} N/m^{2}

Initial volume (V_{1}) = 1 liter = 1 dm^{3} = 1 x 10^{-3} m^{3}

Final volume (V_{2}) = 3 liters = 3 dm^{3} = 3 x 10^{-3} m^{3 }

**Isobaric process CD **:

Pressure (P) = 4 atm = 4 x 10^{5} N/m^{2}

Initial volume (V_{1}) = 2 liters = 2 dm^{3} = 2 x 10^{-3} m^{3}

Final volume

_{2}) = 5 liters = 5 dm

^{3}= 5 x 10

^{-3}m

^{3}

__Wanted ____:__ Difference of the work is done by the gas in process AB and CD.

__Solution :__

Work is done by the gas in process AB :

W = P (V_{2} – V_{1})

W = (6 x 10^{5})(3 x 10^{-3 }– 1 x 10^{-3})

W = (6 x 10^{5})(2 x 10^{-3})

W = 12 x 10^{2 }= 1200 Joule

Work is done by the gas in process CD :

W = P (V_{2} – V_{1})

W = (4 x 10^{5})(5 x 10^{-3 }– 2 x 10^{-3})

W = (4 x 10^{5})(3 x 10^{-3})

W = 12 x 10^{2 }= 1200 Joule

Difference of the work is done by the gas in process AB and CD = 1200 – 1200 = 0.

3. Work is done by the gas in process ABC is….

__Known :__

Pressure 1 (P_{1}) = 6 x 10^{5} Pa = 6 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 cm^{3} = 2 x 10^{-6} m^{3}

Volume 2 (V_{2}) = 6 cm^{3} = 6 x 10^{-6} m^{3}

__Wanted__ : Work is done in process ABC.

__Solution :__

In process AB, the volume is kept constant so that no work is done by the gas.

Work was done by the gas in the process BC.

W = P_{2} (V_{2} – V_{1})

W = (3 x 10^{5})(6 x 10^{-6} – 2 x 10^{-6})

W = (3 x 10^{5})(4 x 10^{-6})

W = 12 x 10^{-1}

W = 1.2 Joule

Work is done in the process ABC = work is done in the process AB = 1.2 Joule.