Physics Help Forum Changing state of an ideal gas

 Thermodynamics and Fluid Mechanics Thermodynamics and Fluid Mechanics Physics Help Forum

 Aug 17th 2016, 02:56 PM #1 Junior Member   Join Date: Aug 2016 Posts: 4 Changing state of an ideal gas Hello everyone, I've been given following question: "An ideal gas is enclosed in a container at state (p1,V1,T1). It is desired to change the state of the gas to (2p1,V1,2T1). This has to be done by only two processes (i.e isothermal followed by isobaric) the processes can't be of the same type. The processes are to be chosen between adiabatic, isobaric, isochoric, and isothermal. In how many ways can this be done?" My thinking is that i can do at least the following: Isochoric -> to some temperature and pressure -> adiabatic to the desired state The reverse Isochoric -> to some temperature and pressure -> isothermal to the desired state The reverse Adiabatic -> to the correct temperature curve -> isothermal to the desired state The reverse Leaving me with 6 different ways to solve the problem. Is this right way of thinking? And is my attempt even correct? Thank you very much in advance.
 Aug 17th 2016, 03:03 PM #2 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 What makes you think this is possible?
 Aug 17th 2016, 03:11 PM #3 Junior Member   Join Date: Aug 2016 Posts: 4 Well as p1,V1 and T1 can be chosen arbitrarily together with the ideal gas, I see no reason why this should not be possible?
Aug 18th 2016, 12:30 AM   #4
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 Well as p1,V1 and T1 can be chosen arbitrarily together with the ideal gas
Can they?

Under what circumstances?

Look carefully at the equation of state for an ideal gas.

If the above is to be true what needs to change to also make

2p1, V1 and 2T1 possible?

Aug 18th 2016, 02:05 AM   #5
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 Can they?
Hmm well if the gas is ideal then the following relation should be true:

PV=nRT

with R as the gas constant. "n" the number og moles, and P,V,T as pressure, volume and temperature.

But as for the state (p1,V1,T1), the volume will most likely be preset by the size of the container,
so once that is chosen then choosing either p1 or T1 will restrain the other one.
I mean if one were to chose p1 then the temperature will be given as T = PV/nR,
and choosing T1 the pressure will be given as p =nRT/V.

And for the rest of I can only guess unfortunately.

 Aug 18th 2016, 03:45 AM #6 Senior Member   Join Date: Apr 2015 Location: Somerset, England Posts: 1,035 So you cannot choose P, V and T independently for a given quantity of gas. The only way choose all three is to play with n, the quantity of gas.
 Aug 18th 2016, 03:56 AM #7 Junior Member   Join Date: Aug 2016 Posts: 4 Okay I see, but let's say we fix the quantity of the gas in a closed container, and we then measure the initial state to be (p1,V1,T1). Is it then possible to change the state to (2p1,V1,2T1) at all or is my guess of 6 path way off?
Aug 18th 2016, 04:33 AM   #8
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 If the above is to be true what needs to change to also make 2p1, V1 and 2T1 possible?