# Thermodynamics problem

• subzero0137
In summary: By holding T constant, the value of U is not affected by changes in V, so the partial derivative is 0.

## Homework Statement

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Use the equation $$C_p - C_v = \left[ P + \left( \frac {∂U}{∂V} \right)_T\right] \left[ \left( \frac {∂V}{∂T} \right)_P \right]$$ to find ##C_p - C_v## for an ideal monatomic gas.

## Homework Equations

##U = \frac {3}{2} RT##
##PV = RT##

## The Attempt at a Solution

I substitute ##PV = RT## into the expression for ##U## to get ##U = \frac {3}{2} PV##, therefore

$$\left( \frac {∂U}{∂V} \right)_T = \frac {3}{2} P$$

Since ##PV=RT ⇒ V = \frac {RT}{P}##,

$$\left( \frac {∂V}{∂T} \right)_P = \frac {R}{P}$$

Therefore ##C_p - C_v = [P + \frac {3}{2} P][\frac {R}{P}] = \frac {5}{2} R## but the real answer is ##C_p - C_v = R##. What have I done wrong? I can't seem to find the mistake.

subzero0137 said:
I substitute ##PV = RT## into the expression for ##U## to get ##U = \frac {3}{2} PV##, therefore

$$\left( \frac {∂U}{∂V} \right)_T = \frac {3}{2} P$$
The T suffix in
$$\left( \frac {∂U}{∂V} \right)_T$$
means keeping T constant. In performing the differentiation you appear to have kept P constant instead.

haruspex said:
The T suffix in
$$\left( \frac {∂U}{∂V} \right)_T$$
means keeping T constant. In performing the differentiation you appear to have kept P constant instead.

Oh I see. So the partial derivative would be 0?

subzero0137 said:
Oh I see. So the partial derivative would be 0?
Yes.

## 1. What is thermodynamics?

Thermodynamics is the branch of science that deals with the study of energy and its transformations in various systems, including physical, chemical, and biological systems.

## 2. What is a thermodynamics problem?

A thermodynamics problem is a question or scenario that involves the application of thermodynamic principles to analyze and understand the behavior of energy in a given system.

## 3. What are the laws of thermodynamics?

The laws of thermodynamics are fundamental principles that govern the behavior of energy in a system. They include the first law, which states that energy cannot be created or destroyed, only transferred or converted, and the second law, which states that the total entropy of a closed system will always either remain constant or increase over time.

## 4. How do you solve a thermodynamics problem?

To solve a thermodynamics problem, you need to identify the system in question, determine the relevant variables and parameters, and apply the appropriate laws and equations to analyze the behavior of energy in the system. It often involves using mathematical equations and manipulating variables to find a solution.

## 5. Why is thermodynamics important?

Thermodynamics is important because it provides a fundamental understanding of how energy behaves in various systems, which has numerous practical applications in fields such as engineering, chemistry, and biology. It also helps us understand and predict the behavior of matter and energy in the universe.