The discussion revolves around a homework problem involving the thermodynamic properties of a diatomic perfect gas heated at constant volume. Participants explore calculations for heat (q), work (w), change in internal energy (ΔU), and change in enthalpy (ΔH), while addressing concepts related to heat capacity and isochoric processes.
Participants generally agree that no work is done in this constant volume process and that the relationship between internal energy and heat can be simplified under these conditions. However, there are differing views on the correct formulation of the equations involved, and some uncertainty remains regarding the application of heat capacity.
Some participants express confusion about the role of pressure and volume in the equations, indicating that assumptions may be necessary regarding standard conditions. The discussion also reflects varying levels of understanding about the implications of heat capacity for diatomic gases.
Students studying thermodynamics, particularly those preparing for exams or needing clarification on concepts related to gas behavior at constant volume.
You don't need to know what a diatomic gas is to solve this problem. They give you the value of the heat capacity. That's all you need.Diamond101 said:Homework Statement
A sample consisting of one mole of a diatomic perfect gas is heated from 25 °C to 200 °C at constant volume. Calculate q, w, ΔU and ΔH for the process, given that Cv = 23.02 J K-1 mol-1 .
Homework Equations
Im studying for mid terms and i have no lecture notes on diatomic gas looking online currently haven't found much. Please explain
The Attempt at a Solution
du=w-cvdt
cv=5/2nr
how do i find work , work it -pdv right? i wasnt given pressure or is it understood as 1 bar and i wasnt given a volumeChestermiller said:You don't need to know what a diatomic gas is to solve this problem. They give you the value of the heat capacity. That's all you need.
If the volume of gas is constant, how much work is done?
What is that Cv doing that equation that you wrote? The equation is incorrect. It should read ΔU=Q-W
If the volume doesn't change, what is dv equal to?Diamond101 said:how do i find work , work it -pdv right? i wasnt given pressure or is it understood as 1 bar and i wasnt given a volume
0, and because w=pdv no work is done ? is this an isochoric process? would du=n cvdtChestermiller said:If the volume doesn't change, what is dv equal to?
Very nice. Yes. isochoric is synonymous with constant volume. And your equation for the differential change in internal energy is correct.Diamond101 said:0, and because w=pdv no work is done ? is this an isochoric process? would du=n cvdt
thank you ever so muchChestermiller said:Very nice. Yes. isochoric is synonymous with constant volume. And your equation for the differential change in internal energy is correct.
Chet