Find the change in entropy for an ideal gas undergoing a reversible process

AI Thread Summary
The discussion focuses on calculating the change in entropy for an ideal gas during a reversible process. The initial approach uses the first law of thermodynamics and integrates expressions for internal energy and pressure. Corrections are suggested, emphasizing the need for a general expression for internal energy that accounts for the number of moles and molar heat capacity. The participants confirm the importance of including the factor of the number of moles in the calculations. Overall, the conversation highlights the complexities of thermodynamic calculations and encourages further practice.
mcas
Messages
22
Reaction score
5
Homework Statement
An ideal gad had a temperature ##T_1## and volume ##V_1##. As a result of a reversible process, these quantities changed to ##T_2## and ##V_2##. Find the change in entropy.
Relevant Equations
##pV=nRT##
##\delta Q = TdS##
##dU = \delta Q + \delta W##
##U = \frac{3}{2}kT##
We know that
$$dU=\delta Q + \delta W$$
$$dU = TdS - pdV$$
So from this:
$$dS = \frac{1}{T}dU + \frac{1}{T}pdV \ (*)$$
For an ideal gas:
$$dU = \frac{3}{2}nkdT$$
Plugging that into (*) and also from ##p=\frac{nRT}{V}## we get:
$$S = \frac{3}{2}nk \int^{T_2}_{T_1} \frac{1}{T}dT + R\int^{V_2}_{V_1} \frac{1}{V}dV$$

And so on...

Is this the correct approach to solve this problem? I'm not really sure because I'm still new to thermodynamics.
 
Physics news on Phys.org
Your approach looks good. A couple of things, though.

mcas said:
##U = \frac{3}{2}kT##
This equation is for a monatomic ideal gas and it's missing a factor of ##N## (the number of molecules). But the question does not specify that the gas is monatomic. So, you'll need a more general expression for ##U## (usually expressed in terms of the number of moles ##n##, the molar heat capacity at constant volume, ##C_V##, and ##T##).

$$S = \frac{3}{2}nk \int^{T_2}_{T_1} \frac{1}{T}dT + R\int^{V_2}_{V_1} \frac{1}{V}dV$$
The first term should be corrected according to the remarks above. The second term is missing a factor. Can you spot it?
 
  • Like
Likes etotheipi and mcas
TSny said:
So, you'll need a more general expression for ##U## (usually expressed in terms of the number of moles ##n##, the molar heat capacity at constant volume, ##C_V##, and ##T##).
Ok, thank you. I think I know which one :smile:

TSny said:
The first term should be corrected according to the remarks above. The second term is missing a factor. Can you spot it?
I missed ##n##, right?

Thank you, this means very much! Now I have the motivation to do more problems 😁
 
  • Like
Likes etotheipi and TSny
mcas said:
I missed ##n##, right?
Yes.
 
Thread 'Minimum mass of a block'
Here we know that if block B is going to move up or just be at the verge of moving up ##Mg \sin \theta ## will act downwards and maximum static friction will act downwards ## \mu Mg \cos \theta ## Now what im confused by is how will we know " how quickly" block B reaches its maximum static friction value without any numbers, the suggested solution says that when block A is at its maximum extension, then block B will start to move up but with a certain set of values couldn't block A reach...
TL;DR Summary: Find Electric field due to charges between 2 parallel infinite planes using Gauss law at any point Here's the diagram. We have a uniform p (rho) density of charges between 2 infinite planes in the cartesian coordinates system. I used a cube of thickness a that spans from z=-a/2 to z=a/2 as a Gaussian surface, each side of the cube has area A. I know that the field depends only on z since there is translational invariance in x and y directions because the planes are...
Thread 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
Back
Top