Gas leaking from a steel tank at changing temperature and pressure

[unscientific]

Homework Statement

A steel container contains 300g of ideal gas at a pressure of 1.35 x 10⁶ Pa and temperature of 77^oC. When the tank is checked later, the temperature is dropped to 22^oC and the pressure has fallen to 8.7 x 10⁵ Pa. How many grams of gas leaked out of the tank?

Homework Equations

PV = NkT

The Attempt at a Solution

first, converting: 77^oC = 350.15K and 22^oC = 295.15K

The problem is, i do not know how to link mass to the equation of PV = NkT

 

[Hootenanny]

Hint: Can you relate N - the number of particles to the mass?

 

[unscientific]

Number of particles = Total mass / mass per molecule?

But i do not know the mass per molecule...

 

[unscientific]

I really need an answer quick, my holidays are ending soon and this is my holiday assignment question that i only stuck with..

 

[aniketp]

HINT 2 : Avogardro number...

 

[unscientific]

I tried it again, closest i could get was,


PV = nRT

n = M_total / M_permole = M_total/(mN_A)

Where m = mass per molecule..

 

[eys_physics]

Hey

Assume that each particle have the mass $$m_0$$. Can you now relate the mass to the number of particles N?
Use this relations to express N in terms of m. Now you can relate the mass after $$m_2$$ to the mass before $$m_1$$ and this relation is independent of $$m_0$$ and only depends on known parameters.
Hopefully this can help you to solve the problem.

 

[unscientific]

Ok i solved it! :D:D The crux of this problem is that, the volume before and after do not change, and i managed to equate mass/molecule.

PV = NkT, PV = (M_total/m_o)kT

After the reaction,

P'V' = N'kT', P'V' = (M'_total/m_o)kT

equating m_o = m_o,

I get

M_totalT/V = M'_totalT'/V'

So i solved for M', giving 229.36 g. So, the mass that leaked = (300 - 229.3) g = 70.6g

I forgot that V was constant, that was perplexing me throughout!

 

[Hootenanny]

Good job