## owlet one year ago I need a step-by-step guidance on how to solve this question. I keep getting it wrong. Question below with my solution. I don't think my solution is right though

1. owlet

A gas is confined to a container with a massless piston at the top. A massless wire is attached to the piston. When an external pressure of 2.03 bar is applied to the wire, the gas compresses from 6.40 to 3.20 L . When the external pressure is increased to 2.53 bar, the gas further compresses from 3.20 to 2.56 L . In a separate experiment with the same initial conditions, a pressure of 2.53 bar was applied to the gas, decreasing its volume from 6.40 to 2.56 L in one step. If the final temperature was the same for both processes, what is the difference between q for the two-step process and q for the one-step process in joules?

2. owlet

@Photon336

3. Photon336

@Cuanchi @Rushwr

4. Photon336

@abb0t

5. Rushwr

Q = U + W right?

6. owlet

no U is the energy, q is the heat..so it should be U=q+w I'm not really sure what to do with the one-step process

7. Rushwr

Bubs what I meant was that is the standard equation right?

8. Photon336

@owlet but if U = Q+W then the change in U deltaU should be equal to the change in Q + change in W. based on the way you've done it implies that the change in work is equal to the -of the Q so that means that the change in internal energy would be zero.

9. Photon336

I think @Woodward @Empty would be able to help solve this question

10. Rushwr

delta Q = delta u + delta W delta Q= heat supplied delta U = increase in internal energy Delta W = work done That is a standard equation. Infact that is the first the law of thermodynamics.

11. Rushwr

This is an isothermal process right? So delta U = 0 Then delta Q = delta W right?

12. Rushwr

that means the only thing we have to do is finding delta W right?

13. owlet

yeah, difference between work done of the one-step process and the two-step process

14. Rushwr

yp

15. Photon336

iSO = SAME thermal temperature. so the temperature is constant.

16. owlet

wait.. so it means.. delta w= w3- (w1+w2) ? w3 is the one step process and (w1 + w2) is the two step process

17. Photon336

Maybe this could help us understand this: the derivation of work at constant pressure. work = pdv isothermal. pV = nRT $\frac{ nRT }{ V } = P$ substitute this into P $p \int\limits_{v_f}^{v_i} dv$ $\int\limits_{v_f}^{v_i} \frac{ nRT }{ v } dV = nRT \int\limits_{V_f}^{V_I} \frac{ 1 }{ v } dV$ $\int\limits_{v_f}^{v_i} \frac{ nRT }{ v } dV = nRT \int\limits_{V_f}^{V_I} \frac{ 1 }{ v } dV = nRTLn \frac{ v_{2} }{ v_1 }$

18. Photon336

this must be an isothermal process Work = nRT*Ln(v2)-Ln(v1) for isothermal expansion if this is an isothermal process the the internal energy change is 0 $\Delta U = 0$ $\Delta U = 0 = q + w ;$ that would mean that q = -w $q = -w$

19. Rushwr

Wait I'm tryna attach my work can u please check on that @Photon336 @owlet

20. owlet

i already got 160J which is the same as the answer on the book. i just did what i understand awhile ago: delta w= w3- (w1+w2)

21. Rushwr

Hold on I can't attach it

22. Rushwr

Oh then that's fine ! I think I might have gone wrong somewhere I got 1600J LOL

23. owlet

maybe you got wrong with converting 1 Lbar : 100J thanks btw for both of your help :)