. Consider a non-ideal gas with the equation of state:
where b is a constant.
i) Derive expressions for: (P/T)V and (T/V)P
ii) For a constant temperature process involving this non-ideal gas, show that
(E/V)T=0
(HINT: Start with an expression for dE, and note that one of the Maxwell relations:
(P/T)V = (S/V)T might be helpful).
iii) What does (E/V)T mean physically, and explain WHY this gas exhibits this special property.
iv) What is (E/V)T for an ideal gas
The printed symbols are Greek on my computer. I can't read them.
i) To derive the expressions for (δP/δT)V and (δT/δV)P, we start with the given equation of state:
P = (RT / V - b) - a / V^2
To find (δP/δT)V, we take the partial derivative of P with respect to T while keeping V constant:
(δP/δT)V = (∂P/∂T)V = (∂/∂T)((RT / V - b) - a / V^2)
Differentiating the equation above, we get:
(δP/δT)V = R / V
To find (δT/δV)P, we take the partial derivative of T with respect to V while keeping P constant:
(δT/δV)P = (∂T/∂V)P
To find this, we can make use of one of the Maxwell relations:
(δP/δT)V = (δS/δV)T
Rearranging the equation, we have:
(δT/δV)P = (δS/δV)T * (δT/δP)V
Now substituting (δP/δT)V = R / V from the previous result, we have:
(δT/δV)P = (δS/δV)T * (V / R)
ii) Now we need to show that (δE/δV)T = 0 for a constant temperature process involving this non-ideal gas.
The first law of thermodynamics states:
dE = δQ - δW
Where dE is the change in internal energy, δQ is the heat added to the system, and δW is the work done by the system.
For a constant temperature process (dT=0), the change in internal energy (dE) is zero since the internal energy only depends on temperature.
Therefore, δQ = δW
To find out (δE/δV)T, we focus on the work done term (δW) which is given by:
δW = -PdV
Where P is the pressure and dV is the change in volume.
Using the previous equation of state, we can substitute P with the equation and rearrange to find dV:
dV = - (δW / P) = (1 / P)(∂E / ∂V)T
Note that (∂E / ∂V)T is (δE / δV)T.
Substituting this back into the first law of thermodynamics equation:
δQ = (1 / P)(∂E / ∂V)T
Since δQ = δW, we have (∂E / ∂V)T = 0.
iii) (δE/δV)T physically means the rate of change of internal energy with respect to volume at constant temperature. It tells us how the internal energy of the system changes as the volume is changed while keeping the temperature constant. For this non-ideal gas, (δE/δV)T = 0 means that the internal energy does not change with volume when the temperature is held constant. This special property can be explained by the fact that this non-ideal gas follows a specific equation of state, which results in the cancellation of the rate of change of internal energy with respect to volume at constant temperature.
iv) For an ideal gas, the equation of state is given by the ideal gas law:
P = nRT / V
Taking the derivative of this equation with respect to V at constant temperature:
(δP/δV)T = -(nRT / V^2)
Substituting this into the equation for (δE/δV)T in part ii:
(δE/δV)T = 1 / P * (δW/δV)T = -1 / P * (δP/δV)T
= -1 / P * (-(nRT / V^2))
Simplifying, we have:
(δE/δV)T = nR / V
Therefore, for an ideal gas, (δE/δV)T is not zero, and it is given by (δE/δV)T = nR / V.