Calculating Density of Helium Gas at Different Temperatures

[laker88116]

The density of helium gas at 0.0 degrees C is 0.179 kg/m^3. The temperature is then raised to 100.0 degrees C, but the pressure remains constant. Assuming that helium is an ideal gas, calculate the new density of the gas.

Ok, so far i figure that since P is constant, I would need to work with v1/t1=v2/t2. I converted temperatures into kelvin (273 and 373 respectively), so I have gotten to v1/273k=v2/373k. How do I find volume for the first or second situation, and how do I use it to find density?

Thanks, David

 

[Doc Al]

Here's a hint: If the volume doubled, how would that affect the density? (Now figure out by what factor the volume changes in this case.)

 

[laker88116]

well since density is mass divided by volume, assuming that mass is constant, (which i think it is since the premise is that its an ideal gas), they are inversely related so like if volume is doubled, density is halved, but how does that help?

 

[laker88116]

ok i did v1/v2=t1/t2, then i substituted 22.4/p1 for v1 and 22.4/p2 for v2, since p=m/v and v=m/p, then i said (22.4/.179)/(22.4/p2)=273/373, and i solved for p2 to get .173 kg/m^3, is that how its done?

 

[dextercioby]

It does,because you can use the relation:[tex] V=\frac{m}{\rho} [/tex]...U need to plug this relation in the gas law...For each of the 2 cases/values of volume.

Daniel.

 

[dextercioby]

ok i did v1/v2=t1/t2, then i substituted 22.4/p1 for v1 and 22.4/p2 for v2, since p=m/v and v=m/p, then i said (22.4/.179)/(22.4/p2)=273/373, and i solved for p2 to get .173 kg/m^3, is that how its done?

No,the problem has no reference to any value of pressure,nor volume...So you can't use any of those numbers...

Daniel.

 

[laker88116]

i meant p as rho for density

 

[laker88116]

hmm I am not seeing this at all

 

[dextercioby]

"p"is for pressure...[tex] \frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}} [/tex] (1)

[tex] V_{1}=\frac{m}{\rho_{1}}[/tex] (2)

[tex] V_{2}=\frac{m}{\rho_{2}} [/tex](3)

Use these three relations to get a connection between [itex] \rho_{1} [/itex] and [itex] \rho_{2} [/itex]...

Daniel.

 

[laker88116]

v=1/p but i said that, i said that they were inversely related if mass is constant

 

[Doc Al]

ok i did v1/v2=t1/t2, then i substituted 22.4/p1 for v1 and 22.4/p2 for v2, since p=m/v and v=m/p, then i said (22.4/.179)/(22.4/p2)=273/373, and i solved for p2 to get .173 kg/m^3, is that how its done?

[itex]\rho_2 = (V_1/V_2) \rho_1 = (T_1/T_2) \rho_1 [/itex]

 

[laker88116]

[itex]\rho_2 = (V_1/V_2) \rho_1 = (T_1/T_2) \rho_1 [/itex]

Thats how i did it though, it yields the same answer.

 

[Doc Al]

Thats how i did it though, it yields the same answer.

If you did what I suggested, you would not end up with 0.173 kg/m^3 as the new density. (Not sure what you did, but your answer is incorrect.)

 

[laker88116]

hmm what did u get then?

 

[Doc Al]

hmm what did u get then?

Take the formula I gave in post #11 and plug in the numbers.

 

[laker88116]

density=(273)/(373) * .179 = .131

 

[Doc Al]

density=(273)/(373) * .179 = .131

Right. (Assuming proper units.)