Proof Ideal Gas: (dU/dV)T=0 & (dH/dP)T=0

In summary, for an ideal gas, the internal energy per unit mass (U) and enthalpy per unit mass (H) are only functions of temperature (T). Therefore, both (dU/dV)T and (dH/dP)T are equal to zero, as volume (V) and pressure (P) are not independent variables for an ideal gas. This can be derived by substituting the ideal gas equation into the expressions for U and H. These equations have been found in a thermodynamics textbook and are not related to any homework question.
  • #1
Hong1111
4
0
How to prove that

(a)(dU/dV)T=0
(b)(dH/dP)T=0

for an ideal gas.

Where U is internal energy per unit mass, V is volume, T is temperature (which is held constant for above 2 question), H is the enthalpy per unit mass, and P is the pressure.

I found this in a thermodynamics textbook. This is not a homework question. I just want to know how does this apply to ideal gas and how to derive the both equations. And in the end, I got warning from moderator.

Thanks.
 
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  • #2
For an ideal gas you should find that the internal energy is only a function of T, hence dU/dV is definitely zero.

Same applies for the enthalpy, define H = U + PV, use the ideal gas equation to substitute for PV and you'll see that H is only a function of T as well, so dH/dP is zero.

I don't see the point of those equations though =/
 

Related to Proof Ideal Gas: (dU/dV)T=0 & (dH/dP)T=0

1. What is the ideal gas law and how does it relate to the proof of (dU/dV)T=0 and (dH/dP)T=0?

The ideal gas law is a mathematical relationship between the pressure, volume, temperature, and number of moles of an ideal gas. It can be expressed as PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. This law is used to derive the proof of (dU/dV)T=0 and (dH/dP)T=0, which state that at constant temperature, the internal energy and enthalpy of an ideal gas do not change with respect to changes in volume and pressure, respectively.

2. How is the proof of (dU/dV)T=0 and (dH/dP)T=0 derived?

The proof of (dU/dV)T=0 and (dH/dP)T=0 is derived using the first and second laws of thermodynamics, along with the ideal gas law. By applying these laws and simplifying the equations, it can be shown that the internal energy and enthalpy of an ideal gas are independent of changes in volume and pressure at constant temperature.

3. What are the implications of (dU/dV)T=0 and (dH/dP)T=0 for ideal gases?

The implications of (dU/dV)T=0 and (dH/dP)T=0 for ideal gases are that at constant temperature, an ideal gas does not experience any change in internal energy when its volume changes, and it does not experience any change in enthalpy when its pressure changes. This means that the internal energy and enthalpy of an ideal gas are solely dependent on its temperature, and not on its volume or pressure.

4. Is the proof of (dU/dV)T=0 and (dH/dP)T=0 applicable to real gases?

No, the proof of (dU/dV)T=0 and (dH/dP)T=0 is only applicable to ideal gases. Real gases do not follow the ideal gas law perfectly and may experience changes in internal energy and enthalpy with changes in volume and pressure, even at constant temperature.

5. How does the proof of (dU/dV)T=0 and (dH/dP)T=0 impact the behavior of ideal gases?

The proof of (dU/dV)T=0 and (dH/dP)T=0 suggests that ideal gases behave in a very predictable manner at constant temperature. This means that their internal energy and enthalpy are not affected by changes in volume and pressure, and their behavior can be easily described using the ideal gas law. It also allows for simplification of calculations involving ideal gases, making them easier to study and understand.

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