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owlet
 one year ago
I need a stepbystep 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
owlet
 one year ago
I need a stepbystep 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

This Question is Closed

owlet
 one year ago
Best ResponseYou've already chosen the best response.1A 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 twostep process and q for the onestep process in joules?

owlet
 one year ago
Best ResponseYou've already chosen the best response.1no 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 onestep process

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.4Bubs what I meant was that is the standard equation right?

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0@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.

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0I think @Woodward @Empty would be able to help solve this question

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.4delta 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
 one year ago
Best ResponseYou've already chosen the best response.4This is an isothermal process right? So delta U = 0 Then delta Q = delta W right?

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.4that means the only thing we have to do is finding delta W right?

owlet
 one year ago
Best ResponseYou've already chosen the best response.1yeah, difference between work done of the onestep process and the twostep process

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0iSO = SAME thermal temperature. so the temperature is constant.

owlet
 one year ago
Best ResponseYou've already chosen the best response.1wait.. so it means.. delta w= w3 (w1+w2) ? w3 is the one step process and (w1 + w2) is the two step process

Photon336
 one year ago
Best ResponseYou've already chosen the best response.0Maybe 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
 one year ago
Best ResponseYou've already chosen the best response.0this 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
 one year ago
Best ResponseYou've already chosen the best response.4Wait I'm tryna attach my work can u please check on that @Photon336 @owlet

owlet
 one year ago
Best ResponseYou've already chosen the best response.1i 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
 one year ago
Best ResponseYou've already chosen the best response.4Hold on I can't attach it

Rushwr
 one year ago
Best ResponseYou've already chosen the best response.4Oh then that's fine ! I think I might have gone wrong somewhere I got 1600J LOL

owlet
 one year ago
Best ResponseYou've already chosen the best response.1maybe you got wrong with converting 1 Lbar : 100J thanks btw for both of your help :)
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