# Kinetic theory of gas and first law of thermodynamics – problems and solutions

1. Ideal gases are in a container with a volume of 4 liters and its pressure is 3 atm (1 atm = 105 N.m-2). The ideal gases heated at a constant pressure from 27oC to 87oC. The heat capacity of the gas is 9 J.K-1. What is the final volume of gases and the change of internal energy of gases?

Solution

Isobaric process (constant pressure)

Known :

The initial volume of gas (V1) = 4 liters

The initial temperature of gas (T1) = 27oC + 273 = 300 K

The final temperature of gas (T2) = 87oC + 273 = 360 K

The pressure of gas (P) = 3 atm = 3 x 105 N.m-2

The heat capacity of gas (C) = 9 J.K-1

Wanted : The final volume of gas (V2) and the change of internal energy of gas (ΔU)

Solution :

Determine the final volume using the equation of Charles’s law (isobaric process or constant pressure) : The change in volume :

1 liter = 0.001 m3

Initial volume (V1) = 4 (0.001 m3) = 0.004 m3

Final volume (V2) = 4.8 (0.001 m3) = 0.0048 m3

The change in volume (ΔV) = V2 – V1 = 0.0048 m3 – 0.004 m3 = 0.008 m3.

The change in temperature :

The change in temperature (ΔT) = T2 – T1 = 360 K – 300 K = 60 K

Determine the change of the internal energy (ΔU) of the ideal gas using the equation of the first law of thermodynamics.

ΔU = Q – W

ΔU = the change of the internal energy, Q = heat, W = work

Determine work (W) at constant pressure :

W = P ΔV = (3 x 105)(0.0008) = (3 x 101)(8) = (30)(8) = 240 Joule

Determine heat (Q) using the equation of the heat capacity (C) :

C = Q / ΔT

Q = (C)(ΔT) = (9)(60) = 540 Joule

Determine the change of the internal energy :

ΔU = Q – W = 540 Joule – 240 Joule = 300 Joule.

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2. 6 liters of ideal gases at 2 atm are in a container (1 atm = 105 N.m-2). The gas heated from 27oC to 77oC at a constant pressure. If the heat capacity of gas is 5 J.K-1, what is the final volume and the change of internal energy of the gas.

Solution :

Isobaric process (constant pressure)

Known :

The initial volume of the ideal gases (V1) = 6 liters

The initial temperature of the ideal gases (T1) = 27oC + 273 = 300 K

The final temperature of the ideal gases (T2) = 77oC + 273 = 350 K

The pressure of the ideal gases (P) = 2 atm = 2 x 105 N.m-2

The heat capacity of gases (C) = 5 J.K-1

Wanted: The final volume of the gas (V2) and the change of internal energy of the gas (ΔU)

Solution :

Determine the final volume of gas using the equation of Charles’s law (isobaric process or constant pressure) : The change in volume :

1 liter = 0.001 m3

The initial volume (V1) = 6 (0.001 m3) = 0.006 m3

The final volume (V2) = 7 (0.001 m3) = 0.007 m3

The change in volume (ΔV) = V2 – V1 = 0.007 m3 – 0.006 m3 = 0.001 m3

The change in temperature :

The change in temperature (ΔT) = T2 – T1 = 350 K – 300 K = 50 K

Determine the change of the internal energy (ΔU) of the ideal gases using the equation of the first law of thermodynamics.

ΔU = Q – W

ΔU = the change in the internal energy, Q = heat, W = work

Determine work (W) at constant pressure :

W = P ΔV = (2 x 105)(0.001) = (2 x 102)(1) = (200)(1) = 200 Joule

Determine heat (Q) using the equation of the heat capacity (C) :

C = Q / ΔT

Q = (C)(ΔT) = (5)(50) = 250 Joule

Determine the change in the internal energy :

ΔU = Q – W = 250 Joule – 200 Joule = 50 Joule.

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