Ideal gas - monatomic or diatomic?

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SUMMARY

The discussion focuses on determining whether an ideal gas is monatomic or diatomic using the molar heat capacities at constant pressure (Cp) and constant volume (Cv). Given 5 moles of gas heated from 300K to 500K at constant pressure, the heat transferred is 29.1 kJ. The calculated Cp is 145.5 J K-1, leading to the conclusion that the gas is diatomic, as indicated by the ratio Cp/nR being 3.5, which aligns with the expected value for diatomic gases.

PREREQUISITES
  • Understanding of the ideal gas law (PV = nRT)
  • Knowledge of heat transfer equations (C = dQ/dT)
  • Familiarity with molar heat capacities (Cp and Cv)
  • Concept of degrees of freedom for gases (f = 3 for monatomic, f = 5 for diatomic)
NEXT STEPS
  • Learn about the derivation of the ideal gas law (PV = nRT)
  • Study the relationship between heat capacities (Cp and Cv) for different gas types
  • Explore the concept of degrees of freedom in thermodynamics
  • Investigate real gas behavior and deviations from ideal gas laws
USEFUL FOR

Students studying thermodynamics, physics enthusiasts, and anyone seeking to understand the properties of gases and their heat capacities.

shyguy79
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Homework Statement



5 moles of an idea gas at 300K at a pressure of 1.00 x 10^5 Pa is heated to 500K at constant pressure. The amount of heat transferred is 29.1kJ.

Determine whether a gas is monatomic or diatomic through consideration of the values of the molar heat capacity at constant pressure C sub c,m and at C sub v,m

Homework Equations


PV = nRT
C = dQ/dT
Cp = Cv + nR
Cv = f/2 nR where f = 3 for monatomic gas and f = 5 for diatomic gas

The Attempt at a Solution


No idea where to start... help please!
 
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shyguy79 said:

Homework Statement



5 moles of an idea gas at 300K at a pressure of 1.00 x 10^5 Pa is heated to 500K at constant pressure. The amount of heat transferred is 29.1kJ.

Determine whether a gas is monatomic or diatomic through consideration of the values of the molar heat capacity at constant pressure C sub c,m and at C sub v,m

Homework Equations


PV = nRT
C = dQ/dT
Cp = Cv + nR
Cv = f/2 nR where f = 3 for monatomic gas and f = 5 for diatomic gas

The Attempt at a Solution


No idea where to start... help please!

Hi shyguy79! :smile:

In a constant pressure process you have Cp=dQ/dT.
Can you calculate Cp from that?
What would you get if you divide Cp by nR?
 
Where dQ = 29.1x10^3 and dT = 200K then Cp = 145.5 J K^-1

Is nR = 5 x 8.314 = 41.57 ? but why would you do this?

so Cp/nr = 3.5 ? How does this relate?
 
Yep. That is correct.

What would Cp/nR be for a monatomic gas?
And for a diatomic gas?
 
Cp = (f/2 +1) nR so then Cp/nR = (f/2 +1)

so for a monatomic gas... 3/2 + 1 = 2.5
for a diatomic gas... 5/2 +1 = 3.5

so the gas is diatomic?
 
Right. :)
 
Thank you so much! I've been stuck on this for ages!
 
Just one question though what equation would be referenced for dividing Cp by nR?
 
Not sure what you mean...?

You've used 3 of your relevant equations and solved for "f".
 
  • #10
doh! Yeah, just noticed - thanks again!
 

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