Question

Calculate the internal energy of 1 gm of O2 at NTP and also find the Cp and Cv.

Answer 1

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Answer 2

The heat capacity of anything tells us how much heat is required to raise a certain amount of it by one degree. For a gas we can define a molar heat capacity C - the heat required to increase the temperature of 1 mole of the gas by 1 K.

Q = nCΔT

The value of the heat capacity depends on whether the heat is added at constant volume, constant pressure, etc. Instead of defining a whole set of molar heat capacities, let's focus on CV, the heat capacity at constant volume, and CP, the heat capacity at constant pressure.

Heat Capacity at Constant Volume

Q = nCVΔT

For an ideal gas, applying the First Law of Thermodynamics tells us that heat is also equal to:

Q = ΔEint + W, although W = 0 at constant volume.

For a monatomic ideal gas we showed that ΔEint = (3/2)nRΔT

Comparing our two equations

Q = nCVΔT and Q = (3/2)nRΔT

we see that, for a monatomic ideal gas:

CV = (3/2)R

For diatomic and polyatomic ideal gases we get:

diatomic: CV = (5/2)R

polyatomic: CV = 3R

This is from the extra 2 or 3 contributions to the internal energy from rotations.

Because Q = ΔEint when the volume is constant, the change in internal energy can always be written:

ΔEint = n CV ΔT

Heat Capacity at Constant Pressure

For an ideal gas at constant pressure, it takes more heat to achieve the same temperature change than it does at constant volume. At constant volume all the heat added goes into raising the temperature. At constant pressure some of the heat goes to doing work.

Q = nCPΔT

For an ideal gas, applying the First Law of Thermodynamics tells us that heat is also equal to:

Q = ΔEint + W

At constant pressure W = PΔV = nRΔT

For a monatomic ideal gas, where ΔEint = (3/2)nRΔT, we get:

Q = (3/2)nRΔT + nRΔT = (5/2)nRΔT

So, for a monatomic ideal gas:

CP = (5/2)R

For diatomic and polyatomic ideal gases we get:

diaatomic: CP = (7/2)R

polyatomic: CP = 4R