Need help calculating the pressure in the bottle

[mhn745]

Liquid helium at 4.2 K has a density of 0.147 g/mL. Suppose that a 2.50-L metal bottle that contains air at 105K and 2.0 atm pressure is sealed off. If we inject 100.0 mL of liquid helium and allow the entire system to warm to room temperature (25 °C), what is the pressure inside the bottle?

This is what I have so far
For air alone-
P/T = P1/T1
2.0 / 105 = P1 / 298
P1 = 5.676 atm

For He alone-
100.0 mL x 0.147 g/mL = 1.47 g
1.47 g/ 4.00 g/mol = .3675 mol
P=nRT/V=.3675x0.08206x298/2.50=3.595 atm

Ptotal = 3.595 + 5.7 = 9.296 atm


But apparently the answers wrong. Can someone please help?

Addtional Hints:

At the end of the problem, the bottle contains both helium and air, and we want the total pressure. You know V (same as the initial V of the metal bottle) and T, so if you knew n you could solve for P. Find the number of moles of air and of helium separately, then add them to get the total number of moles of gas in the bottle. I will add that you need to use two pretty different ways to find the moles of the two gases.

does anyone know how to solve for the mole (n) in pv=nrt for this equation...maybe if I find the number of moles individually for both then I can solve...

 

[Redbelly98]

> "100.0 mL x 0.147 g/mL = 1.47 g"

Try that one again.

 

[mhn745]

okay i realized that and fixed it but i still get the wrong answer...i used 14.7 g this time

 

[Redbelly98]

Huh, I'm not sure what else is wrong. But I didn't check the rest of it, did you use consistent units for the ideal gas equation?

 

[mhn745]

yes I used k for temperature and the right constant value

 

[GCT]

Liquid helium at 4.2 K has a density of 0.147 g/mL. Suppose that a 2.50-L metal bottle that contains air at 105K and 2.0 atm pressure is sealed off. If we inject 100.0 mL of liquid helium and allow the entire system to warm to room temperature (25 °C), what is the pressure inside the bottle?

This is what I have so far
For air alone-
P/T = P1/T1
2.0 / 105 = P1 / 298
P1 = 5.676 atm

For He alone-
100.0 mL x 0.147 g/mL = 1.47 g
1.47 g/ 4.00 g/mol = .3675 mol
P=nRT/V=.3675x0.08206x298/2.50=3.595 atm

Ptotal = 3.595 + 5.7 = 9.296 atm


But apparently the answers wrong. Can someone please help?

Addtional Hints:

At the end of the problem, the bottle contains both helium and air, and we want the total pressure. You know V (same as the initial V of the metal bottle) and T, so if you knew n you could solve for P. Find the number of moles of air and of helium separately, then add them to get the total number of moles of gas in the bottle. I will add that you need to use two pretty different ways to find the moles of the two gases.

does anyone know how to solve for the mole (n) in pv=nrt for this equation...maybe if I find the number of moles individually for both then I can solve...

Check your MW for Helium gas.

 

[Borek]

okay i realized that and fixed it but i still get the wrong answer...i used 14.7 g this time

What was your final result?

How many significant digits have you entered?

4.00 is a correct MW for helium.

 

[GCT]

Liquid helium at 4.2 K has a density of 0.147 g/mL. Suppose that a 2.50-L metal bottle that contains air at 105K and 2.0 atm pressure is sealed off. If we inject 100.0 mL of liquid helium and allow the entire system to warm to room temperature (25 °C), what is the pressure inside the bottle?

This is what I have so far
For air alone-
P/T = P1/T1
2.0 / 105 = P1 / 298
P1 = 5.676 atm

For He alone-
100.0 mL x 0.147 g/mL = 1.47 g
1.47 g/ 4.00 g/mol = .3675 mol
P=nRT/V=.3675x0.08206x298/2.50=3.595 atm

Ptotal = 3.595 + 5.7 = 9.296 atm


But apparently the answers wrong. Can someone please help?

Addtional Hints:

At the end of the problem, the bottle contains both helium and air, and we want the total pressure. You know V (same as the initial V of the metal bottle) and T, so if you knew n you could solve for P. Find the number of moles of air and of helium separately, then add them to get the total number of moles of gas in the bottle. I will add that you need to use two pretty different ways to find the moles of the two gases.

does anyone know how to solve for the mole (n) in pv=nrt for this equation...maybe if I find the number of moles individually for both then I can solve...

Is the answer 42 atm?