Help with this Law of Thermodynamics question please

[Gusk]

MODERATOR'S NOTE: NO TEMPLATE BECAUSE THREAD WAS ORIGINALLY POSTED IN A NON-HOMEWORK FORUM
A monoatomic gas is present i n cylinder if its volume increase from 100cm3 to 200cm3 at constant pressure 1*10^5 what is change in internal energy. I tried to solve it but couldn't.
I tried to solve it by using first law of thermodynamics.
∆U+W=Q
W=P∆V
The only thing I needed was Q but how to get it?
I thought maybe Q=C∆T but T is not given so a bit confused how to solve.

 

[kuruman]

How about showing us what you tried and we'll take it from there. It would be appropriate to attach units to the pressure you quoted.

 

[Gusk]

How about showing us what you tried and we'll take it from there. It would be appropriate to attach units to the pressure you quoted.

I have edited

 

[Gusk]

Although it is not homework and it is just a question to prepare for admission in university but I would like hints or something's so that I learn something new and solve it myself.

 

[berkeman]

Although it is not homework and it is just a question to prepare for admission in university

Welcome to the PF. We treat all schoolwork-type questions (even for self-study and review) the same, so they all go here in the Homework Help forums, and use the Template that is provided when you start a thread here.

 

[Chestermiller]

Although it is not homework and it is just a question to prepare for admission in university but I would like hints or something's so that I learn something new and solve it myself.

How much work is done by the gas on its surroundings?

 

[kuruman]

How much work is done by the gas on its surroundings?

We still don't know the units of pressure so we can't figure that out ... yet.

 

[Gusk]

MODERATOR'S NOTE: NO TEMPLATE BECAUSE THREAD WAS ORIGINALLY POSTED IN A NON-HOMEWORK FORUM
A monoatomic gas is present i n cylinder if its volume increase from 100cm3 to 200cm3 at constant pressure 1*10^5 pa what is change in internal energy. I tried to solve it but couldn't.
I tried to solve it by using first law of thermodynamics.
∆U+W=Q
W=P∆V
The only thing I needed was Q but how to get it?
I thought maybe Q=C∆T but T is not given so a bit confused how to solve.

 

[Gusk]

We still don't know the units of pressure so we can't figure that out ... yet.

Sorry, its pascal.

 

[Chestermiller]

So, algebraically, what is the equation for the work done by the gas on its surroundings?

 

[Gusk]

So, algebraically, what is the equation for the work done by the gas on its surroundings?

I couldn't understand what you mean?
Maybe W=P∆V

 

[Chestermiller]

I couldn't understand what you mean?
Maybe W=P∆V

OK. Now, plug that into the first law of thermodynamics. What do you get?

 

[Gusk]

OK. Now, plug that into the first law of thermodynamics. What do you get?

1*10^7 J

 

[kuruman]

1*10^7 J

That's not what I get. Can you list separately what you got for ΔW and ΔQ?

On edit: I assumed there is one mole of the gas.

 

[Chestermiller]

Is the cylinder supposed to be insulated (I.e., adiabatic)?

 

[kuruman]

Is the cylinder supposed to be insulated (I.e., adiabatic)?

Can there be adiabatic expansion at constant pressure?

 

[Chestermiller]

Can there be adiabatic expansion at constant pressure?

Sure...irreversible expansion at constant externally applied pressure.

However, as you are seem to be indicating (and I agree), the original intent of this problem was probably to consider the expansion as reversible, in which case the expansion would not be adiabatic. There would have to be heat transfer to keep the gas pressure at the original value. Sorry if I may have confused the issue.

 

[Gusk]

That's not what I get. Can you list separately what you got for ΔW and ΔQ?

On edit: I assumed there is one mole of the gas.

W = P∆V
= ( 1*10^5)(200-100)=1*10^7 J

 

[Gusk]

Is the cylinder supposed to be insulated (I.e., adiabatic)?

I simply posted the whole question so I don't know.

 

[Gusk]

That's not what I get. Can you list separately what you got for ΔW and ΔQ?

On edit: I assumed there is one mole of the gas.

And I don't know how to find ∆Q. Maybe nC∆T and I don't have ∆T.

 

[kuruman]

And I don't know how to find ∆Q. Maybe nC∆T and I don't have ∆T.

You have the ideal gas law.

 

[Gusk]

You have the ideal gas law.

I see so P∆V/nR = T right?

 

[kuruman]

I see so P∆V/nR = T right?

P∆V/nR = ΔT.

 

[Gusk]

P∆V/nR = ΔT.

If put values in it I get ∆T= 1.2K then how do I find Cp in Q=nCp∆T?

 

[Chestermiller]

If put values in it I get ∆T= 1.2K then how do I find Cp in Q=nCp∆T?

$$\Delta U=nC_v\Delta T=nC_v\left(\frac{P\Delta V}{nR}\right)=\frac{C_v}{R}(P\Delta V)$$
$$Q=\Delta U+P\Delta V$$

 

[Gusk]

$$\Delta U=nC_v\Delta T=nC_v\left(\frac{P\Delta V}{nR}\right)=\frac{C_v}{R}(P\Delta V)$$
$$Q=\Delta U+P\Delta V$$

We have to find ∆U and answer is 150 J and how do I find Cv?

 

[Chestermiller]

We have to find ∆U and answer is 150 J and how do I find Cv?

You need to go back to your textbook. It's right in there. Your knowledge base is in serious need of review.

 

[Dr Dr news]

The first thing you need to do is use a consistent set of units. kg, m. sec with the derived unit for force, Newton, n = kg -m/sec^2. You are looking for the change in internal energy, ΔU = n Cv ΔT. Since you are considering a monatomic gas, Cv = (3/2)R. The next move is to derive a function for ΔT. From the perfect gas law pV = nRT, T = pV/nR; Ti = pi Vi /nR, Tf = pf Vf /nR, ΔT = p ΔV/nR, ΔU = n Cv ΔT = n(3/2)R*pΔV/nR = (3/2)pΔV = (3/2)10^5 n/m^2 *10^-4 m^3 = 15 J. I think you were very close but once again I would emphasize that you need to use a consistent set of units and label every number in your calculations with the proper units. I've been in this business for a long time and this is the major difference between an A student and one which ends up recycling.