# Calculate the Enthropy Change of Isothermal Expansion of a Non-Ideal Gas

• Chemistry
• Ignitia
In summary, the conversation discusses the calculation of ΔS using the formula ΔS = ∫ (dq/T) and the given values of dq and T. In part (a), the formula is simplified to ΔS = nR[ln(V2/V1)]. In part (b), the conversation talks about solving for v using the equation [P + a/(v/n)2][v/n - b] = RT and using the Maxwell states to find dS/dV. Finally, in part (c), the conversation uses the previously calculated value for ΔS to solve for the final answer.
Ignitia
Homework Statement
a) Calculate the entropy change for an isothermal expansion of a Van Der Waals Gas from V1 to V2. b) Use this to calculate ΔS for 1 mole of NH3 from 2L to 20L at 298K. c) Compare this to an Ideal Gas
Relevant Equations
ΔS = ∫ (dq/T)
PV=nRT Ideal Gas
[P + a/(v/n)[SUP]2[/SUP]][v/n - b] = RT Not Ideal
Part (a)
ΔS = ∫ (dq/T)

because: dq = PdV = (nRT/V)dV

Then:
ΔS = ∫ (1/T)*(nRT/V)dV
ΔS = nR ∫(1/V) dV
ΔS =nR[ln(V2/V1)]

Part (b)
This is where I'm stuck. I know [P + a/(v/n)2][v/n - b] = RT can be solved for P and simplified to
P = [RT/(v-b)]-[a/v2] since n=1mol

But I don't know how to proceed from here, to solve for v and in turn solve ΔS. I can't use Boyle's Law and I was told there's another method besides Newton-Rapson.

Part (c)
Take ΔS =nR(ln(V2/V1) from part (a) and input: V2=20L |V1=2L

ΔS = (1mol)(8.3145 J*mol/K)(ln(10))=19.15 J/K

19.15J/K - (part b) = Final Answer

I think I got it:
P = [RT/(v-b)]-[a/v2]

Integrate with respect to T:

dP/dT = RT/[(v/n)-b] (d/dT) - [a/v2] (d/dT)

[a/v2] (d/dT) goes to 0 so

dP/dT = R/[(v/n)-b]

And a Maxwell states: dP/dT = dS/dV

so: dS/dV = R/[(v/n)-b]

dS = R/[(v/n)-b] dV

ΔS = ∫ R/[(v/n)-b] dV

And then just integrate from 2L to 20L? Is this correct for Part b?

## What is the definition of entropy change?

Entropy change is a measure of the amount of disorder or randomness in a system. It is also known as the increase in entropy or entropy production.

## How is entropy change calculated for isothermal expansion of a non-ideal gas?

The entropy change for isothermal expansion of a non-ideal gas can be calculated using the equation: ΔS = nR ln(V₂/V₁), where n is the number of moles, R is the gas constant, V₂ is the final volume, and V₁ is the initial volume.

## What is the role of temperature in calculating entropy change?

Temperature is a crucial factor in determining the entropy change of a system. In isothermal expansion, the temperature remains constant, which means that there is no change in the internal energy of the gas. This results in the entropy change being directly proportional to the change in volume.

## How does the behavior of a non-ideal gas affect the entropy change of isothermal expansion?

A non-ideal gas does not follow the ideal gas law, which means that its behavior is affected by intermolecular forces and the volume occupied by the gas particles. This results in a deviation from the ideal gas behavior, leading to a different value for entropy change compared to an ideal gas under the same conditions.

## What are some factors that can influence the entropy change of isothermal expansion of a non-ideal gas?

Some factors that can affect the entropy change of isothermal expansion of a non-ideal gas include the type of gas, initial and final volumes, and the temperature at which the expansion occurs. Additionally, the presence of impurities, such as other gases or particles, can also influence the entropy change.

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