First Law of Thermodynamics

The first law can be represented as

    \[ U = Q + W \]

U is the internal energy of the system. This is defined exactly by the extrinsic variables of the system. For our purposes we can assume the internal is only a function of temperature where U = U(T). This is true for an ideal gas. W is the work done on the system and Q is the heat transferred to the system.

The equation of state for an ideal gas is

    \[ PV = nRT \]

This is very useful in defining how changes can happen in a hydrostatic system.

The differential form of the first law is

    \[ dU = dQ + dW \]

If U is a function of the thermodynamic coordinates of the system T, P and V then we have

    \[ dU_P = \left( \frac {\partial U} {\partial T} \right)_V dT + \left ( \frac{\partial U}{\partial V} \right )_T dV \]

and

    \[ dU_V = \left( \frac {\partial U} {\partial T} \right)_P dT + \left ( \frac{\partial U}{\partial P} \right )_T dP \]

But as U=U(T) then

    \[ \left ( \frac{\partial U}{\partial V} \right )_T = 0 \]

and

    \[ \left ( \frac{\partial U}{\partial P} \right )_T = 0 \]

For a hydrostatic system, the work done on the system is given by

    \[ dW = -PdV \]

as the change in volume implies compression is a negative change when the fluid is compressed.

Therefore we can rewrite the first law as

    \[ dQ = dU + PdV \]

So for an isochoric change, V is constant, we have

    \[ dQ_V = \left ( \frac{\partial U}{\partial T} \right )_V dT +  \left ( \frac{\partial U}{\partial P} \right )_T dP \]

or

    \[ dQ_V = \left ( \frac{\partial U}{\partial T} \right )_V dT \]

Giving

    \[ \left ( \frac{dQ}{dT} \right )_V = \left ( \frac{\partial U}{\partial T} \right )_V = \frac{dU}{dT} = C_V \]

where C_V is the specific heat capacity of the fluid at constant volume.

So for an isobaric change, P is constant, we have

    \[ dQ_P = \left ( \frac{\partial U}{\partial T} \right )_P dT +  \left ( \frac{\partial U}{\partial V} \right )_T dV + PdV \]

so

    \[ dQ_P = C_V dT + PdV \]

differentiating with respect to T

    \[ \left ( \frac{dQ}{dT} \right )_P = C_V + P \frac{dV}{dT} \]

where

    \[ \left ( \frac{dV}{dT} \right ) = \frac{nR}{P} \]

so

    \[ C_P = C_V + nR \]

For molar specific heat capacities

    \[ c_P = c_V + R \]

This is a definite observable result. Most gases behave like ideal gases at low pressure and high temperature (room temperature). So looking at observable measurements of c_p and c_v for known gases we find

for oxygen c_p = 29.4 and c_v = 21.1 so c_p - c_v = 8.33 and \frac{c_p}{c_v} = 1.40