How Does Temperature Affect the Density of Helium Gas?

[BunDa4Th]

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.

 

[Andrew Mason]

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

 

[BunDa4Th]

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?

 

[Andrew Mason]

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