Solve Adiabatic Process of Ideal Monatomic Gas (γ=5/3)

[cdeggz]

A fixed quantity of an ideal monatomic gas (γ = 5/3) is being taken around a cycle

p1 = 10 atm
V1 = 2 L
T1 = 370 K
V2 = V3 = 15 L
p2= 4/3

Here's what's happening: During the isothermal process, the gas expands while in contact with a large heat reservoir, so its temperature remains constant (T2 = T1).

During the constant volume (isochoric) process, the gas is cooled from T2 to T3 by putting it in contact with a reservoir at temperature T3. (This is an irreversible process.)
During the adiabatic process, the gas is compressed while isolated from outside heat sources, so no heat flows. Work is done on the gas to compress it. The temperature rises from T3 to T1, so the gas has returned to its original state.

What is the pressure of state 3?


-I can't get this. I have p1T1^(5/3) = p3T3^(5/3), but that is 1 equation with 2 variables. What else can I uses? I don't know the work in this phase either.

 

[siddharth]

Lets us say that the gas is first at P1,V1,T1

After the first stage (Isothermal expansion) the gas goes from P1,V1,T1 to P2,V2,T2.
Not because the process is isothermal. T1=T2. Since P1,V1,V2 are given P2 is found

During the isochoric process, the gas goes from P2,V2,T2 to P3,V3,T3.

Because the process is isochoric V2=V3. Since T2 and T3 are given P3 can be found.

Now remember that in an adiabatic process PV^gamma is constant.( NOT PT^gamma)
Because you know all other variables, the final pressure can be found

Note: while approaching this problem you have directly jumped from step 1 to step 3.
A better approach would be to write down the state of the gas in each step and then relate the variable with the given data. This bit of extra effort will ensure that you do not go wrong or get confused in the middle of the problem.

 

[bullados]

The problem is that T3 isn't a given quantity. T3 must be found out. Here is the problem in its entirety...

(We're probably both in the same class, phys 213 at UofI, so it's the same problem)

A fixed quantity of an ideal monatomic gas (γ = 5/3) is being taken around the cycle shown in this p-V diagram.

p1 = 10 atm
V1 = 2 L
T1 = 370 K
V2 = V3 = 15 L

Here's what's happening:
During the isothermal process, the gas expands while in contact with a large heat reservoir, so its temperature remains constant (T2 = T1).
During the constant volume (isochoric) process, the gas is cooled from T2 to T3 by putting it in contact with a reservoir at temperature T3. (This is an irreversible process.)

During the adiabatic process, the gas is compressed while isolated from outside heat sources, so no heat flows. Work is done on the gas to compress it. The temperature rises from T3 to T1, so the gas has returned to its original state.


a) What are the state 2 and 3 pressures?

The state 2 pressure is 4/3 atm. How can you find T3?

[EDIT] Never mind. Found it. It has to do with the VT^alpha constant equations on the adiabatic side. [/EDIT]

 

[ZapperZ]

Are you aware that you're tackling a problem that was asked in Nov. 2004? I can safely guess that the boat has left the harbor already a long time ago.

Zz

 

[bullados]

Nope, this problem was asked this week. I didn't notice the date when putting this problem into Google looking for an answer. Sorry! The school loves recycling old questions...