Physics stuff | East London Science School

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

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

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

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

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

$dQ_P = C_V dT + PdV$

differentiating with respect to T

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

where

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

$C_P = C_V + nR$

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 $\frac{c_p}{c_v} = 1.40$