Thermo: isothermal, reversible expansion of ideal gas

In summary, the gas expands reversibly and isothermally to twice its original volume, doing work along the way.
  • #1
Ryaners
50
2

Homework Statement


Two moles of a monatomic ideal gas are at a temperature of 300K. The gas expands reversibly and isothermally to twice its original volume. Calculate the work done by the gas, the heat supplied and the change in internal energy.
So:
T = 300K; ΔT = 0
n = 2; R = 8.314 J K-1 mol-1
V2 = 2V1
Reversible process ⇒ work is path independent.

Homework Equations


PV = nRT
W = -PdV
U = W + Q

The Attempt at a Solution


PV = nRT ⇒ P = nRT / V
dW = -PdV ⇒ W = - ∫ PdV = - nRT ∫ [dV / V] = nRT [ln |V2 - V1|] = nRT ln V1

This isn't very helpful as I have no actual figure for what V1 is.

I've tried to come up with something based on how the work is path independent (so W = P2V2 - P1V1 = V1[P2 - P1]) but I've just confused myself, as ΔT = 0 means that P1V1 = P2V2 = nRT, so that would mean that W = 0 which isn't right. I must be going wrong with my assumptions somewhere - any pointers would be much appreciated!

Edit: I'm confident that once I get past the first stumbling block & calculate the work done I can fire ahead & calculate Q and ΔU. Thanks in advance!
 
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  • #2
You integrated incorrectly. Try again.
 
  • #3
Chestermiller said:
You integrated incorrectly. Try again.

Aha! I see what you mean & I get the right answer when evaluating the integral like this:
nRT [ln |V2 - V1|] = nRT [ln |V2 / V1| ] = nRT [ln | 2V1 / V1| ] = nRT ln[2]

Thanks for the nudge in the right direction. May I ask: is it ever correct to write
nRT [ln |V2 - V1|]
or should it be
nRT [ln |V2| - ln |V1|]
when substituting end points in an integral of this form? Are these two expressions equivalent? (I'm now thinking they're not..?)
 
  • #4
Ryaners said:
Aha! I see what you mean & I get the right answer when evaluating the integral like this:
nRT [ln |V2 - V1|] = nRT [ln |V2 / V1| ] = nRT [ln | 2V1 / V1| ] = nRT ln[2]

Thanks for the nudge in the right direction. May I ask: is it ever correct to write
nRT [ln |V2 - V1|]
or should it be
nRT [ln |V2| - ln |V1|]
when substituting end points in an integral of this form? Are these two expressions equivalent? (I'm now thinking they're not..?)
They're not equivalent. You need to review some of the math related to this. When you evaluate results of a definite integral at the limits of integration, you write ##[f(x)]_{x=a}^{x=b}=f(b)-f(a)##, not ##f(b-a)##. Also, the argument of a combined natural log term should have no units.
 
  • #5
Chestermiller said:
They're not equivalent. You need to review some of the math related to this. When you evaluate results of a definite integral at the limits of integration, you write ##[f(x)]_{x=a}^{x=b}=f(b)-f(a)##, not ##f(b-a)##. Also, the argument of a combined natural log term should have no units.

Thanks, that makes it very clear. It's more that I have some uncertainty around logarithms, which I need to revise / practice again; i.e. ln(a) - ln(b) ≠ ln(a-b), which is obvious to me now but was the mistake I made yesterday without questioning.

Appreciate the help, thank you!
 

What is an isothermal expansion?

An isothermal expansion is a process in which the temperature of a system remains constant while the volume of the system increases. This means that the internal energy of the system also remains constant.

What is a reversible expansion?

A reversible expansion is a process in which the system can be returned to its original state by reversing the direction of the process without any changes to the surroundings. This means that the system reaches equilibrium at every step of the process.

What is an ideal gas?

An ideal gas is a hypothetical gas that follows the ideal gas law, which states that the pressure, volume, and temperature of the gas are all directly proportional. In other words, the behavior of an ideal gas can be described by a simple mathematical relationship.

Why is the expansion of an ideal gas considered to be isothermal and reversible?

The expansion of an ideal gas is considered to be isothermal because the temperature of the gas remains constant throughout the process. It is also considered to be reversible because the gas can be returned to its original state by reversing the process without any changes to the surroundings.

What are the applications of isothermal, reversible expansion of ideal gas in science?

The isothermal, reversible expansion of ideal gas is a common process used in thermodynamics, and it has many practical applications in science. Some examples include heat engines, refrigeration systems, and gas turbines.

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