Stat Mech- How does heat vary with temperature over an arbitrary pathway

[mandir08]

Homework Statement

You are told that you have 1 mole of an ideal gas with heat capacity at constant volume being 1.5R and you send it over an arbitrary path where dq/dt|pathway= 2R. In the end, the volume of the gas doubles, so figure out by what factor the temperature must change. Assume that the process is reversible

The Attempt at a Solution

du=dq +dw= dq -p*dv
dq=du + p*dv
dq/dt|path =du/dt |path + (d/dt* (P)* dv + P dv/dt|path)

Saying that P= -dF/dv|t,n cause it to become

dq/dt|path =du/dt |path + (0+ P dv/dt|path)

dq/dt|path=2R

u=Cv*T=NRC *T (I am not sure I can apply equation of state u=Cv*T
du/dt=Cv=1.5R =NRC

2R =du/dt |path + (0+ P dv/dt|path)
2R =1.5R + (0+ P dv/dt|path)=RC + (0+ P dv/dt)
.5R=P* dv/dt|path
P/.5R=dt/dv|path

V*P/.5R=T such that T2=V2*P2/.5R=2V1*P2/.5R

However because I am sure that I did this wrong because I can get the same result knowing that PV=NRT such that T2=2*(P2/P1) T1

The hint at the end about the process being reversible makes it that Cv=(du/dt)|v=T(ds/dt)|v

I am kind of lost at this point and not really sure how to proceed. Anyone want to help point me in the right direction?

 

[mark726]

it's important to approach problems like this with a systematic and logical approach. Let's break down the information given and see what we can deduce from it.

1. We have 1 mole of an ideal gas.
2. The heat capacity at constant volume (Cv) is 1.5R.
3. The heat flow rate along the pathway (dq/dt|pathway) is 2R.
4. The volume of the gas doubles.
5. The process is reversible.

First, let's recall the definition of heat capacity at constant volume (Cv). It is the amount of heat required to raise the temperature of 1 mole of a substance by 1 Kelvin at constant volume. So in this case, we know that the heat flow rate (dq/dt) is equal to 2R. This means that for every second, 2R amount of heat is flowing into our system.

Next, let's consider the fact that the gas is undergoing a reversible process. This means that the gas is always in equilibrium with its surroundings, and the system is always in a state of thermodynamic equilibrium. This is important because it allows us to use the equation of state for an ideal gas, PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is the temperature.

Now, let's think about what happens to the gas as its volume doubles. We know that the number of moles (n) and the gas constant (R) are constant, so the only variable that can change is the temperature (T). Using the ideal gas law, we can set up the following equation:

(P1)(V1) = (P2)(V2)

where P1 and V1 are the initial pressure and volume, and P2 and V2 are the final pressure and volume. We also know that the final volume (V2) is twice the initial volume (V1), so we can rewrite the equation as:

(P1)(V1) = (P2)(2V1)

Next, we can rearrange the equation to solve for the final pressure (P2):

P2 = (P1)(V1)/(2V1) = P1/2

Since we know that the pressure and number of moles are constant, we can also use the ideal gas law to solve for the final temperature (