# 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 = 10^{5} N.m^{-2}). The ideal gases heated at a constant pressure from 27^{o}C to 87^{o}C. 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 (V_{1}) = 4 liters

The initial temperature of gas (T_{1}) = 27^{o}C + 273 = 300 K

The final temperature of gas (T_{2}) = 87^{o}C + 273 = 360 K

The pressure of gas (P) = 3 atm = 3 x 10^{5} N.m^{-2}

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

__Wanted :__ The final volume of gas (V_{2}) 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 m^{3 }

Initial volume (V_{1}) = 4 (0.001 m^{3}) = 0.004 m^{3}

Final volume (V_{2}) = 4.8 (0.001 m^{3}) = 0.0048 m^{3}

The change in volume (ΔV) = V_{2} – V_{1 }= 0.0048 m^{3} – 0.004 m^{3} = 0.008 m^{3}.

__The change in temper ature :__

The change in temperature (ΔT) = T_{2} – T_{1} = 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 10^{5})(0.0008) = (3 x 10^{1})(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.

2. 6 liters of ideal gases at 2 atm are in a container (1 atm = 10^{5} N.m^{-2}). The gas heated from 27^{o}C to 77^{o}C 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 (V_{1}) = 6 liters

The initial temperature of the ideal gases (T_{1}) = 27^{o}C + 273 = 300 K

The final temperature of the ideal gases (T_{2}) = 77^{o}C + 273 = 350 K

The pressure of the ideal gases (P) = 2 atm = 2 x 10^{5} N.m^{-2}

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

__Wanted:__ The final volume of the gas (V_{2}) 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 m^{3 }

The initial volume (V_{1}) = 6 (0.001 m^{3}) = 0.006 m^{3}

The final volume (V_{2}) = 7 (0.001 m^{3}) = 0.007 m^{3}

The change in volume (ΔV) = V_{2} – V_{1} = 0.007 m^{3} – 0.006 m^{3} = 0.001 m^{3}

__The change in temperature :__

The change in temperature (ΔT) = T_{2} – T_{1 }= 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 10^{5})(0.001) = (2 x 10^{2})(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.