Question in Thermal Physics (Van der Waals' Equation)

[warhammer]

Homework Statement

Calculate the critical temperature for a gas obeying van der Waals equation of state. Given a = 0.00874 atm cm^6 and b = 0.0023 cm³ for 1 cm³ of the gas at S.T.P.

Ans: T = 307 K

(Hint: First calculate the values of a and b for 1 mole of the gas.)

Relevant Equations

Critical Temperature Tc= (8a)/(27Rb)

<Using the hint, I tried to find the van der Waal constants in molar form. Since STP is mentioned, I used the unitary method relationship-
22.4 L=22400cm^3=1 molar V

<To find a possible conversion standard between cm^3 and mol; which turned out to be 1cm^3= 4.46*10^-5 mol.

<Then I used the above attained value and used them wrt 'a' & 'b', in order to convert their units from atm*cm^6 & atm*cm^3 to atm*mol^2 and atm*mol respectively.

However I'm not getting the correct answer. Please guide me where I am making the error.

(Have also attached a photo of my solution attempt)

 

[mjc123]

How do a and b scale with V? If you change V from 1 cm³ to 22400 cm³, should these constants get bigger or smaller?

 

[warhammer]

How do a and b scale with V? If you change V from 1 cm³ to 22400 cm³, should these constants get bigger or smaller?

The constants would get bigger once we scale them for 22400 cm^3 since right now they account for only 1cm^3 of gas

 

[Steve4Physics]

First consider b, as it’s easy to understand its meaning (excluded volume per amount of gas) and see what’s happening.

The unit for b is typically cm³.mol⁻¹ (volume per amount of gas).

But the question gives the value for b in a very confusing and peculiar (to me) unit. Rather than being specified in ‘cm³ per mole’ it is specified in units of 'cm³ per [however many moles there are in 1cm³ of the gas at STP]’.

For the purpose of this explanation, let’s call [however many moles there are in 1cm³ of the gas at STP] a ‘Z’. So we are told b = 0.0023 cm³.Z⁻¹.

If we treat the gas as ideal, 1 mole occupies 22400cm³ at STP, so we have
1 Z = ¹/₂₂₄₀₀ mol
giving
1 Z⁻¹ = 22400 mol⁻¹

b = 0.0023 cm³.Z⁻¹ = 0.0023 cm³ x 22400 mol⁻¹ = 51.52 cm³.mol⁻¹

A similar argument can be used to convert the given value of ‘a’ from atm.cm⁶.Z⁻² to atm.cm⁶.mol⁻²

If you then use R in appropriate units, this leads to $T_c$ = 307 K.
________________

A couple of other thoughts…

‘22.4 litres ≡ 1 mole at STP’ is true for an ideal gas (a=0, b=0), but not for a real gas. However, in this problem, it is a sensible approximation – and using it gives the required final answer of 307K.

Your detailed working is not always shown, so the origin(s) of any error(s) will not be clear. E.g. we can’t tell the value & unit you used for R.

 

[warhammer]

First consider b, as it’s easy to understand its meaning (excluded volume per amount of gas) and see what’s happening.

The unit for b is typically cm³.mol⁻¹ (volume per amount of gas).

But the question gives the value for b in a very confusing and peculiar (to me) unit. Rather than being specified in ‘cm³ per mole’ it is specified in units of 'cm³ per [however many moles there are in 1cm³ of the gas at STP]’.

For the purpose of this explanation, let’s call [however many moles there are in 1cm³ of the gas at STP] a ‘Z’. So we are told b = 0.0023 cm³.Z⁻¹.

If we treat the gas as ideal, 1 mole occupies 22400cm³ at STP, so we have
1 Z = ¹/₂₂₄₀₀ mol
giving
1 Z⁻¹ = 22400 mol⁻¹

b = 0.0023 cm³.Z⁻¹ = 0.0023 cm³ x 22400 mol⁻¹ = 51.52 cm³.mol⁻¹

A similar argument can be used to convert the given value of ‘a’ from atm.cm⁶.Z⁻² to atm.cm⁶.mol⁻²

If you then use R in appropriate units, this leads to $T_c$ = 307 K.
________________

A couple of other thoughts…

‘22.4 litres ≡ 1 mole at STP’ is true for an ideal gas (a=0, b=0), but not for a real gas. However, in this problem, it is a sensible approximation – and using it gives the required final answer of 307K.

Your detailed working is not always shown, so the origin(s) of any error(s) will not be clear. E.g. we can’t tell the value & unit you used for R.

Ah! Through your valuable insights, I finally deduced where I was making a very silly error (Yes the wording of the question seemed a bit odd to me too).

Certainly, I will keep your guidance in mind about approximating a real gas into an ideal gas situation. Very grateful for your support Steve, thank you loads!

 

[Quiser]

Ah! Through your valuable insights, I finally deduced where I was making a very silly error (Yes the wording of the question seemed a bit odd to me too).

Certainly, I will keep your guidance in mind about approximating a real gas into an ideal gas situation. Very grateful for your support Steve, thank you loads!

Can you please share the detailed solution of this problem. Today I also face with same problem and can't able to figure out logic.
So please send me solution.

Thanking you..!

 

[berkeman]

Can you please share the detailed solution of this problem. Today I also face with same problem and can't able to figure out logic.
So please send me solution.

Thanking you..!

Nope, nope, nope. At PF, you need to do the work on your schoolwork questions. Please start a new thread in the schoolwork/homework forums and show your work on the solution. Then you will get great tutorial help.