Ideal gas law problem

[Jacob87411]

1. A rigid tank contains .5 kmol of Ar and 2kmol of N2 at 250 kPa and 280 K. The mixture is now heated to 400 K. Determine the volume of the tank and the final pressure of the mixture.



2. PV=NRT

Can you find the initial volume by using PV=NRT?

V=NRT/P
V=(2.5)(8.314)(280)/250=23.2 m^3

Now I am confused how I used this to find V2 and P2, is there another relation between pressure and volume that I don't have? Thanks

 

[f(x)]

find partial pressures (mole fraction of each gas x total pressure) and add them up to get total pressure.
Volume is tank is constant

 

[Dick]

find partial pressures (mole fraction of each gas x total pressure) and add them up to get total pressure.
Volume is tank is constant

If you are going to do it that way be sure and compute a 'partial volume' for each gas as well. But why bother? Both pressures go up by the same fraction. Why split them?

 

[Jacob87411]

V1=V2=23.2 m^3 or you mean the volume of the tank is constant because I thought the 23.2 was thevolume of the gas not the tank

 

[Dick]

In your solution to computing the volume of the gas, you set N=2.5 kmol. So you treated the two gasses as one big volume of ideal gas. There is no need to split them into species. A mix of ideal gasses is itself an ideal gas.

 

[Jacob87411]

Yes but I am still confused. How do you use the mole ratio to find the pressure after heating it up

 

[t!m]

No mole ratios necessary. As Dick said, just assume you have 2.5 kmol of one ideal gas, this is N. Using this value for N, you can find the volume of the container (watch your units though!). Now plug this volume into the ideal gas law, along with the new temperature, to determine the new pressure.

 

[f(x)]

If you are going to do it that way be sure and compute a 'partial volume' for each gas as well. But why bother? Both pressures go up by the same fraction. Why split them?

partial volume ?? what i meant was to use dalton's law of partial pressures...which states total pressure is sum of partial pressures of each gas taken separately (but volume remains constant)

initially we have v=nRT/p (calculate with initial conditions)

now that u know the volume(it remains const as the container is rigid)
calculate the partial pressures of each gas for t=400 K and add them up to get the total pressure (P2)
ie. p(Ar)=moles of Ar*R*T / V
p(N2)=moles of N2*R*T / V
p2=p(Ar)+p(N2)

The thing is, since names of gases are explicitly mentioned while nothing about their ideal behaviour is mentioned, i wonder if its correct to use ideal gas eqn.
Check what answer you get with the above else we shall have to use any real gas eqn like vanderwaals eqn.

 

[Dick]

I see what you are saying. That IS the more correct way to state the solution. Thought since the poster had combined the gasses for the first problem, he may as well continue. But I think your presentation is much better. That is, in fact, probably why the two gasses were named - to make you mention partial pressures (not for a nonideality problem). Sorry to muddy the waters.

 

[Gokul43201]

V1=V2=23.2 m^3 or you mean the volume of the tank is constant because I thought the 23.2 was thevolume of the gas not the tank

Jacob: Can the gas occupy a different volume than the tank? Look up the different definitions of 'gas' .

Yes, V1=V2.


f(x): You have used the ideal gas equation in your last post, so you too have assumed both gases are ideal (that seems a reasonable assumption to make, especially given the choices of gases in the question). There is no need to deal with partial pressures.

 

[f(x)]

umm...yeah the scope of the question seems limited to ideal gas eqn at the moment.
oh, lol and sorry for the partial pressures...like Dick said b4...it doesn't make a difference if they are ideal gases