How Does Temperature Affect the Density of Helium Gas?

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SUMMARY

The discussion focuses on calculating the new density of helium gas when the temperature is increased from 0°C to 105°C at constant pressure. The initial density of helium at standard temperature and pressure (STP) is established as 0.179 kg/m³. Using the Ideal Gas Law (PV = nRT), participants derive the volume of one mole of helium and confirm the density calculation. The universal gas constant R is clarified as 8.314 J/(mol·K), with a distinction made regarding its alternative value of 0.0831 in different unit systems.

PREREQUISITES
  • Understanding of the Ideal Gas Law (PV = nRT)
  • Basic knowledge of gas density calculations
  • Familiarity with temperature conversion between Celsius and Kelvin
  • Concept of standard temperature and pressure (STP)
NEXT STEPS
  • Learn how to apply the Ideal Gas Law to different gases under varying conditions
  • Study the implications of temperature and pressure changes on gas density
  • Explore the concept of molar volume at STP for various gases
  • Investigate the differences in gas constant values across different unit systems
USEFUL FOR

Students studying chemistry, particularly those focusing on gas laws, as well as educators and anyone interested in the properties of gases under varying thermal conditions.

BunDa4Th
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Homework Statement



The density of helium gas at T = 0°C and atmospheric pressure is 0 = 0.179 kg/m3. The temperature is then raised to T = 105°C, but the pressure is kept constant. Assuming that helium is an ideal gas, calculate the new density f of the gas. kg/m3

Homework Equations



PV = nRT
m/v = weight in grams?

The Attempt at a Solution



T1 = 0*C T2 = 105*C
Po = .179 Pf = ?
Pa1 = 1 atm? Pa2 = 10 atm?

I think that is how I am suppose to set it up but from there I am at a complete lost on what to do. But i am not even sure where to start on this problem.
 
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BunDa4Th said:

Homework Statement



The density of helium gas at T = 0°C and atmospheric pressure is 0 = 0.179 kg/m3. The temperature is then raised to T = 105°C, but the pressure is kept constant. Assuming that helium is an ideal gas, calculate the new density f of the gas. kg/m3

Homework Equations



PV = nRT
m/v = weight in grams?

The Attempt at a Solution



T1 = 0*C T2 = 105*C
Po = .179 Pf = ?
Pa1 = 1 atm? Pa2 = 10 atm?

I think that is how I am suppose to set it up but from there I am at a complete lost on what to do. But i am not even sure where to start on this problem.
The key here is to find the volume change per unit mass.

They give you the density of He at STP. But let's calculate it. Assume you have one mole of gas. Calculate the volume of one mole at 273K (0 C) and 1 atm (101325 Pa) using the Ideal gas law:

V = nRT/P = 1*8.314*273/101325 = .0224 m^3 = 22.4 L

Since one mole of He has mass of 4g, the density is .004/.0224 = .179 kg/m^3.

Do the same for He at a pressure of 378K (105C) and 10 atm. That is all the question is asking.

AM
 
I have a question on how did you get R to equal to 8.314?

I thought R would be .0831 or am i looking at a different R value?
 
BunDa4Th said:
I have a question on how did you get R to equal to 8.314?

I thought R would be .0831 or am i looking at a different R value?
In MKS:

R = N_A k_B = 8.31451 m^2kg/s^2 K mol

The .0831 must be in units of m^2kg/s^2 K mol \times 10^3 or kilomoles

AM
 
Last edited:

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