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- owlet

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

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- owlet

- jamiebookeater

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- 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?

- owlet

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- Photon336

- Rushwr

Q = U + W right?

- 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

- Rushwr

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

- 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.

- 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.

- Rushwr

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

- Rushwr

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

- owlet

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

- Rushwr

yp

- Photon336

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

- owlet

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

- 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
\[p \int\limits_{v_f}^{v_i} dv \]
\[\int\limits_{v_f}^{v_i} \frac{ nRT }{ v } dV = nRT \int\limits_{V_f}^{V_I} \frac{ 1 }{ v } dV \]
\[\int\limits_{v_f}^{v_i} \frac{ nRT }{ v } dV = nRT \int\limits_{V_f}^{V_I} \frac{ 1 }{ v } dV = nRTLn \frac{ v_{2} }{ v_1 }\]

- 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
\[q = -w \]

- Rushwr

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

- 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)

- Rushwr

Hold on I can't attach it

- Rushwr

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

- owlet

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

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