Air in a piston/cylinder assembly at 200kpa and 600K is expanded in a constant-pressure process to twice the initial volume (state 2). The piston is then locked with a pin, and heat is transferred to a final temperature of 600K. (assume cp and cv are constant).
A) find P, T and h for states 2 and 3
B) find 1W3 and 1Q3.
I need help with B!!! i can't find volume to do the work equation, someone please help

Question

Air in a piston/cylinder assembly at 200kpa and 600K is expanded in a constant-pressure process to twice the initial volume (state 2). The piston is then locked with a pin, and heat is transferred to a final temperature of 600K. (assume cp and cv are constant).
A) find P, T and h for states 2 and 3
B) find 1W3 and 1Q3.
I need help with B!!! i can't find volume to do the work equation, someone please help

kainui · Answer

If you know the information from A, can you use that along with PV=nRT to find the volume to help you in part B?

anonymous · Answer

My issue was that I don't know the value or n or V so I couldn't solve for the mass.

kainui · Answer

$$P_1V_1=nRT_1$$Let's just say this describes the first state. We're given some stuff already, $$P_1=200 \ kPa \ T_1=600 \ K$$ now they're saying that they expanded it to twice the original volume while keeping the pressure the same. That means since it's a closed container, the temperature had to change as well, so let's write it out $$P_2V_2=nRT_2$$ But we can plug in a coupble values, since pressure was constant and volume doubled, we can write that mathematically as $$P_1=P_2 \ V_2=2V_1$$
Now we do a third thing too it by locking the pin and changing the temperature $$P_3V_3=nRT_3$$ Locking the pin keeps the volume the same, and we changed the temperature back to the original temperature. So that means we can rewrite this as $$P_3V_2=nRT_1$$

You can calculate the work without knowing the number of molecules since Work is just dependent on pressure and volume.

anonymous · Answer

so this is where i am at now:
1W3= 1W2+2W3
since volume is constant from process 2 to 3, 2W3=0
therefore:
1W3=1W2
1W2=(P1+P2)*(V2-V1)/2
I know the pressures, but i do not know the volumes
Which equation do you think i can manipulate to find V2 or V1?

anonymous · Answer

YO welcome back Kai

kainui · Answer

Hmm this is interesting, I don't think there's enough information here to solve it or something lol. I thought I got the answer earlier but I think I was mistaken.

kainui · Answer

Can you leave it as a variable? Since V2=2*V1 then that means V2-V1 is simply V1 by itself. Hmm.

anonymous · Answer

While you are here, do you know how to solve this question?

Suppose that X ∼ N(−7, 14). Find:
(h) The value of x for which P(|X + 7| ≤ x) = 0.44