How much oxygen escaped from the container?

In summary: Sorry for the confusion. The initial pressure is 4 atm and the final pressure is 3.5 atm, so the change in pressure is 0.5 atm. The initial mass of oxygen in the container is 341.1 g, and the final mass is 340.6 g. Therefore, the amount of oxygen that escaped is 0.5 g. In summary, the initial mass of oxygen in the container was 341.1 g and the final mass was 340.6 g, resulting in 0.5 g of oxygen escaping due to a decrease in pressure from 4 atm to 3.5 atm.
  • #1
Karol
1,380
22

Homework Statement


Into a 56[liter] container oxigen is filled in with manometric pressure of 4[atm] and temp' 470C. what is the mass.
Then a leak happens and the pressure drops to 3.5[atm] and temp' 270C.
How much oxigen escaped.

Homework Equations


$$PV=nRT$$
$$R\left[\frac{liter\cdot atm'}{mole\cdot ^0K}\right]=0.08208$$
Atomic number O2=16

The Attempt at a Solution


The initial conditions: ##3[atm]\cdot 56[liter]=n\cdot 0.08208\cdot 320^0K\rightarrow n=6.4[mole]##
$$m=6.4\cdot 16=102.3[gr]$$
Second stage:
$$\frac{P_1V_1}{n_1T_1}=\frac{P_2V_2}{n_2T_2}\rightarrow\frac{P_1V_1}{m_1T_1}=\frac{P_2V_2}{m_2T_2}$$
Our volume is fixed, so: ##\frac{3}{102.3\cdot 320}=\frac{2.5}{m_2\cdot 300}\rightarrow m_2=90.97[gr]##
Is it correct?
 
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  • #2
Karol said:

Homework Statement


Into a 56[liter] container oxigen is filled in with manometric pressure of 4[atm] and temp' 470C. what is the mass.
Then a leak happens and the pressure drops to 3.5[atm] and temp' 270C.
How much oxigen escaped.

Homework Equations


$$PV=nRT$$
$$R\left[\frac{liter\cdot atm'}{mole\cdot ^0K}\right]=0.08208$$
Atomic number O2=16

The Attempt at a Solution


The initial conditions: ##3[atm]\cdot 56[liter]=n\cdot 0.08208\cdot 320^0K\rightarrow n=6.4[mole]##
$$m=6.4\cdot 16=102.3[gr]$$

Why did you calculate with 3 atm?
And the oxygen gas consists of O2 molecules. The atomic mass of oxygen is 16 g, but you have to work with the molar mass.
 
  • #3
manometric pressure is the pressure above atmospheric, right? so i have to take absolute pressures, it's then 5 atm.
And the molar mass is 32[gr]
$$5[atm]\cdot 56[liter]=n\cdot 0.08208\cdot 320^0K\rightarrow n=10.7[mole]$$
$$m=10.7\cdot 32=341.1[gr]$$
 
  • #4
Karol said:
manometric pressure is the pressure above atmospheric, right? so i have to take absolute pressures, it's then 5 atm.
And the molar mass is 32[gr]
$$5[atm]\cdot 56[liter]=n\cdot 0.08208\cdot 320^0K\rightarrow n=10.7[mole]$$
$$m=10.7\cdot 32=341.1[gr]$$

I know pressure above atmospheric as "gauge pressure". You have to know what manometric pressure is. I never heard that.
 
  • #5
Yes, it's gauge pressure
 

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