Finding Pressure with constant temperature and varying volume

[joe465]

Homework Statement

A 3-litre vessel contains gas at a pressure of 200 kPa. Five litres of gas at the same pressure is forced into the vessel. Calculate the new pressure of the gas.

Homework Equations

P1V1 = P2V2

The Attempt at a Solution

P1V1 = P2V2 (Keeping Temp = Constant)
V2 = 3 + 5 = 8 litres
3 * 200 = P2 * 8
P2 = 600/8 = 75 kPa


I have been told this answer is wrong but unsure why if anyone could help me out, would greatly appreciate it.

Thanks, Joe

 

[collinsmark]

Homework Statement

A 3-litre vessel contains gas at a pressure of 200 kPa. Five litres of gas at the same pressure is forced into the vessel. Calculate the new pressure of the gas.

Homework Equations

P1V1 = P2V2

The Attempt at a Solution

P1V1 = P2V2 (Keeping Temp = Constant)

Just to be sure, you are working with an ideal gas right?

V2 = 3 + 5 = 8 litres

Are you sure that's the final volume, V₂? It sounds to me from the problems statement that 8 liters is the initial volume, V₁. (Although this volume is split into two separate containers, one 3 liter container and another 5 liter container.)

3 * 200 = P2 * 8
P2 = 600/8 = 75 kPa


I have been told this answer is wrong but unsure why if anyone could help me out, would greatly appreciate it.

If more gas is being forced into the 3 liter container (while keeping temperature the same), you would think the pressure in that container would increase, not decrease, right?

 

[joe465]

Thanks for your reply, well the question has been copied word for word and just says gas so i presume it would be right to assume its an ideal gas.

So you suggest V1 would be 8L and the 5L is forced into the 3L section making V2 3L therefore making the pressure rise.


P1V1 = P2V2 (Keeping Temp = Constant)
V2 = 8 - 5 = 3 litres
8 * 200 = P2 * 3
P2 = 1600/3 = 533.33 kPa

That looks like a more realistic outcome, what do you think?

 

[collinsmark]

P1V1 = P2V2 (Keeping Temp = Constant)
V2 = 8 - 5 = 3 litres
8 * 200 = P2 * 3
P2 = 1600/3 = 533.33 kPa

That looks like a more realistic outcome, what do you think?

'Looks good to me. (This is assuming the temperature is kept constant.)

 

[joe465]

Thankyou very much, it literally only states what it does in the question and doesn't suggest temperature change.

Would the question emply or give a temperature if this was required?

Thanks, Joe

 

[algebrat]

PV=nRT.

Process 1, adding gas at fixed pressure, n changes, V changes, T stays same.

Process 2, with new V, we compress it to old V. Thus in PV=nRT, the pressure and temp are ambiguous. We must compress it in a certain way.

A temp bath would keep the temp constant. A thermos would stop heat exchange (not the same as temp bath). A free piston on top would keep pressure the same.

The problem is ambiguous since it doesn't tell us which process to use for process 2.

That is, either isothermic, adiabatic or isobaric.

 

[collinsmark]

Thankyou very much, it literally only states what it does in the question and doesn't suggest temperature change.

Would the question emply or give a temperature if this was required?

Thanks, Joe

If it explicitly states in the question (such as in the title of this thread) that the temperature remains constant, or if it states that the process is isothermal, then your answer is fine.

Even if it doesn't state anything about temperature, then I'm still guessing that your answer is fine, and the process is assumed to be isothermal.

On the other hand, if the process is adiabatic, then things get a little more complicated. But I doubt this is the case, since the problem statement never mentioned whether the gas was monatomic, diatomic, or whatnot. If your coursework hasn't discussed this stuff (yet), I'd assume that the temperature stays constant, and your answer is correct.