Solve Thermal Physics Homework: Find Mass of Gas Leaked From Tank

AI Thread Summary
A steel container initially holds 300g of an ideal gas at 1.35 x 10^6 Pa and 77°C. After a temperature drop to 22°C, the pressure decreases to 8.7 x 10^5 Pa, prompting the need to calculate the mass of gas that leaked. The solution involves using the ideal gas law, PV = NkT, and relating the number of particles to mass. By equating the mass before and after the leak while keeping volume constant, the final mass of gas is determined to be 229.36g, indicating that 70.6g of gas leaked from the tank. Understanding that the volume remained unchanged was crucial to solving the problem.
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Homework Statement


A steel container contains 300g of ideal gas at a pressure of 1.35 x 106 Pa and temperature of 77oC. When the tank is checked later, the temperature is dropped to 22oC 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: 77oC = 350.15K and 22oC = 295.15K

The problem is, i do not know how to link mass to the equation of PV = NkT
 
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Hint: Can you relate N - the number of particles to the mass?
 
Number of particles = Total mass / mass per molecule?

But i do not know the mass per molecule...
 
I really need an answer quick, my holidays are ending soon and this is my holiday assignment question that i only stuck with..
 
HINT 2 : Avogardro number...
 
I tried it again, closest i could get was,


PV = nRT

n = Mtotal / Mpermole = Mtotal/(mNA)

Where m = mass per molecule..
 
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.
 
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 = (Mtotal/mo)kT

After the reaction,

P'V' = N'kT', P'V' = (M'total/mo)kT

equating mo = mo,

I get

MtotalT/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!
 
Good job :approve:
 

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