Question

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

Answer 1

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?

Answer 2

@Photon336

Answer 3

@Cuanchi @Rushwr

Answer 4

@abb0t

Answer 5

Q = U + W right?

Answer 6

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

Answer 7

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

Answer 8

@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.

Answer 9

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

Answer 10

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.

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

yp

Answer 15

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

Answer 16

wait.. so it means..

delta w= w3- (w1+w2) ?
w3 is the one step process and (w1 + w2) is the two step process

Answer 17

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

\[p \int\limits_{v_f}^{v_i} dv \]

\[\int\limits_{v_f}^{v_i} \frac{ nRT }{ v } dV = nRT \int\limits_{V_f}^{V_I} \frac{ 1 }{ v } dV \]

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

Answer 18

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

\[q = -w \]

Answer 19

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

Answer 20

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)

Answer 21

Hold on I can't attach it

Answer 22

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

Answer 23

maybe you got wrong with converting 1 Lbar : 100J

thanks btw for both of your help :)