Thermodynamic - Work done in a system

[ibysaiyan]

Homework Statement

The question is as following:

One litre of an ideal gas, initially at atmospheric pressure, is held at a constant
temperature of 300 K (by maintaining good thermal contact with a thermal reservoir
held at this temperature). The volume is decreased until the pressure doubles.
Calculate how much work is done on the gas. How much heat flows out of the gas?

Homework Equations

[tex]W = \int P(V) dV[/tex]

The Attempt at a Solution

Now I know that for work done in a varying pressure , I must use the following equation:[tex]W = \int P(V) dV[/tex]
where I sub in ideal gas equation for pressure.

[itex]W = nRT\ln{(V0/V)}[/itex] (since it's compression not expansion)

Information that's given : 1 lite of water = 55.5 moles , T = 300k , Pressure 1 : 1x10^5 , pressure 2 : 2x10^5 , Volume one : 0.001 m3 , volume 2 : 5x10^-4 (since p1v1 = p2v2)After having plugged in the numbers.. I get work of 319 joules .
Am I on the right track ? also since this is an isothermal cycle Internal energy stays constant i.e dE =0, So work done = heat out, right ?

 

[ibysaiyan]

No ,I have just plugged int he same values I get 93 kJoules of work.. is this wrong ?

 

[LawrenceC]

Would you write a line of numbers showing the arithmetic you did in substituting into the formula for the polytropic process.

 

[grzz]

Is the work done 69.3J?

 

[ibysaiyan]

Sure .
We know work = -p(V)dv which when we substitude int the ideal gas equation becomes : W = integral of nrt/v dv 8.31 * 55.5 * 300 * ln(0.001/ 5x10^-4)

Last night I got a value of 320 joules but now when I am plugging the same numbers into my calculator... an absurd quantity shows up.

What am I doing wrong ?

@ grzz: I don't know. hmm

 

[LawrenceC]

Is the work done 69.3J?

That is what I computed.

 

[willem2]

Your equation shouldn't have n, R or T in it. All you need to compute P(V) is [itex] PV = P_0 V_0 [/itex] and you know [itex] P_0 [/itex] and [itex] V_0 [/itex]

 

[grzz]

How did you get the value of the number of moles, n?

 

[ibysaiyan]

That is what I computed.

I can't get that answer... hmm

Your equation shouldn't have n, R or T in it. All you need to compute P(V) is [itex] PV = P_0 V_0 [/itex] and you know [itex] P_0 [/itex] and [itex] V_0 [/itex]

I am confused. The question does give t =300k.

How did you get the value of the number of moles, n?

Well the given quantity of water is 1 litre = 1000g
Moles = mass / mr = 1000 / 18 = 55.5 .

 

[grzz]

PV = nRT
PV/(RT) = n and i do not get 55.5 for n.

 

[ibysaiyan]

PV = nRT
PV/(RT) = n and i do not get 55.5 for n.

What's wrong with my value of n ? I am totally lost on how to solve this question.

 

[grzz]

Well the given quantity of water is 1 litre = 1000g
Moles = mass / mr = 1000 / 18 = 55.5 .

I do not understand why you put 18.

 

[grzz]

The substance is an ideal gas. It is not water.

 

[ibysaiyan]

The substance is an ideal gas. It is not water.

OMG. You're right! there's no mention of water!
In this case would n = 0.040 [itex]W = nRT\ln{(V0/V)}[/itex]

Then let me try to compute in the answer.I FINALLY GOT THE ANSWER OF 69 JOULES!1 YES! YAY.. THANKS EVERYONE! THANKS FOR YOUR HELP..

 

[grzz]

I used to tell my students to THINK and I used to remind them of that very often when correcting their homework.
Let me share this with you. They heard that word so often that once a student turned up with a T-shirt printed 'THINK' on one side and 'E = mc[itex]^{2}[/itex] ' on the other.

 

[ibysaiyan]

I used to tell my students to THINK and I used to remind them of that very often when correcting their homework.
Let me share this with you. They heard that word so often that once a student turned up with a T-shirt printed 'THINK' on one side and 'E = mc[itex]^{2}[/itex] ' on the other.

Thanks for sharing that. It made me chuckle. lol

Such students should often think at a slower rate, there's no of urgency. (I ,myself am included into this category)