Question in Thermal Physics (Van der Waals' Equation)

In summary, the conversation discusses the use of the hint to find the van der Waal constants in molar form, with STP mentioned as a reference point. The unitary method relationship is used to convert 22.4 L to 22400 cm^3 and then to 1 molar V. A conversion standard between cm^3 and mol is found to be 1cm^3= 4.46*10^-5 mol. Next, the value for 'a' and 'b' are converted from units of atm*cm^6 & atm*cm^3 to atm*mol^2 and atm*mol, respectively. However, the correct answer is not obtained and guidance is requested to identify the error
  • #1
warhammer
151
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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)
 

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  • #2
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?
 
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  • #3
mjc123 said:
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
 
  • #4
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.
 
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  • #5
Steve4Physics said:
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! 🙏🏻
 
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  • #6
warhammer said:
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..!
 
  • #7
Quiser said:
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. :smile:
 
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Related to Question in Thermal Physics (Van der Waals' Equation)

1. What is Van der Waals' equation?

Van der Waals' equation is a thermodynamic equation of state that describes the behavior of real gases, taking into account the intermolecular forces and the finite size of gas molecules. It is an improvement over the ideal gas law, which assumes that gas molecules have no volume and do not interact with each other.

2. What are the limitations of Van der Waals' equation?

Van der Waals' equation is not accurate for all gases and conditions. It does not account for the effects of temperature, pressure, and volume on the intermolecular forces, and it does not consider the possibility of phase transitions. It is also less accurate at high pressures and low temperatures.

3. How is Van der Waals' equation derived?

Van der Waals' equation is derived from the kinetic theory of gases and the assumption that gas molecules have a finite size and interact with each other through attractive and repulsive forces. The equation is then modified to include corrections for these intermolecular forces and the volume of the gas molecules.

4. What is the significance of Van der Waals' equation?

Van der Waals' equation is important in understanding the behavior of real gases, which are not ideal. It helps to explain deviations from the ideal gas law and provides a more accurate description of gas behavior under various conditions. It is also used in many practical applications, such as in the design of industrial processes and the development of new materials.

5. How is Van der Waals' equation used in thermodynamics?

Van der Waals' equation is used in thermodynamics to calculate the properties of real gases, such as their volume, pressure, and temperature, under different conditions. It is also used in the study of phase transitions, such as the liquid-gas transition, and in the determination of critical points. It is an important tool in the analysis and prediction of gas behavior in various thermodynamic processes.

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