Clarification on PV=nRT question.

• Rizke
In summary: If the actual atmospheric pressure is known (rather than 100000 Pa), you add that to get the absolute pressure. Otherwise, as an estimate, you use 100000 Pa.
Rizke
Does not use the template because it was originally posted in a non-homework forum
A welder using a tank of volume 7.50×10^−2 m^3 fills it with oxygen (with a molar mass of 32.0 g/mol ) at a gauge pressure of 3.00×10^5 Pa and temperature of 36.5 ∘C. The tank has a small leak, and in time some of the oxygen leaks out. On a day when the temperature is 21.0 ∘C, the gauge pressure of the oxygen in the tank is 1.85×10^5 Pa . Find the initial mass of oxygen. correct answer : .374 kg
I can't seem to get this right. i even worked backwards with that answer and still. I read in a similar thread that i need the absolute value of gas pressure. but not sure how to apply

My Attempt: PV=nRT
P = 3.0e^5 - in pascals
V = 7.5e^-2
R = 8.3144621
T = (36.5 C + 273.15)= 309.65 K
(3.0e^5)(7.5e^-2) / (8.3144621)(309.65) = n => n= 8.7393 moles
mass = 8.7393* 32 = > 279.658 grams. but .280 kg is not .374 kg.
So yeah I am puzzled.

Last edited:
Rizke said:
A welder using a tank of volume 7.50×10^−2 m^3 fills it with oxygen (with a molar mass of 32.0 g/mol ) at a gauge pressure of 3.00×10Pa and temperature of 36.5 ∘C. The tank has a small leak, and in time some of the oxygen leaks out. On a day when the temperature is 21.0 ∘C, the gauge pressure of the oxygen in the tank is 1.85×105 Pa . Find the initial mass of oxygen. correct answer : .374 kg
I can't seem to get this right. i even worked backwards with that answer and still. I read in a similar thread that i need the absolute value of gas pressure. but not sure how to apply

My Attempt: PV=nRT
P = 3.0e^5 - in pascals
V = 7.5e^-2
R = 8.3144621
T = (36.5 C + 273.15)= 309.65 K
(3.0^5)(7.5e^-2) / (8.3144621)(309.65) = n => n= 8.7393 moles
mass = 8.7393* 32 = > 279.658 grams. but .280 kg is not .374 kg.
So yeah I am puzzled.
Do you know the definition of gauge pressure?

oh that's right a gauge will read zero, but there is actually the ATM pressure that is on the gauge. which is like 15 right? so i have to subtract that pressure after i convert it to pascals. or is this not right? Thanks for the quick reply.

Rizke said:
oh that's right a gauge will read zero, but there is actually the ATM pressure that is on the gauge. which is like 15 right? so i have to subtract that pressure after i convert it to pascals. or is this not right? Thanks for the quick reply.
Add 100000 Pa to the gauge pressure to get the absolute pressure.

P = (3.0e^5)+100000 = 4.0e^5
(4.0e^5)(7.5e^-2) / (8.3144621)(309.65) = 11.6524 moles => 11.65*32 = 372.89 grams => .373 kg
Awesome Thank you!
p.s can i always add 100000 Pa for absolute pressure? that is if i already convert ATM to PA.

Rizke said:
P = (3.0e^5)+100000 = 4.0e^5
(4.0e^5)(7.5e^-2) / (8.3144621)(309.65) = 11.6524 moles => 11.65*32 = 372.89 grams => .373 kg
Awesome Thank you!
p.s can i always add 100000 Pa for absolute pressure? that is if i already convert ATM to PA.
If the actual atmospheric pressure is known (rather than 100000 Pa), you add that to get the absolute pressure. Otherwise, as an estimate, you use 100000 Pa.

Okay i think i finally understand why we add to the pressure. Assuming the welder in this problem is at a location where there is one ATM pressure, the gauge does not account for this. So we need to add that 1 ATM, or 101325 PA to the final pressure reading. i get .374 kg

Chestermiller

1. What does the equation PV=nRT represent?

The equation PV=nRT is known as the ideal gas law, where P represents pressure, V represents volume, n represents the number of moles, R is the gas constant, and T is temperature. It describes the relationship between these variables in an ideal gas.

2. What is the value of the gas constant, R?

The gas constant, R, has a value of 0.0821 L·atm/mol·K in SI units. However, it can also be expressed in other units such as J/mol·K or cm^3·atm/mol·K.

3. Can the ideal gas law be applied to all gases?

No, the ideal gas law assumes ideal conditions such as high temperature, low pressure, and no intermolecular forces. This means it is most accurate for gases that behave closely to ideal gases, such as noble gases.

4. How does changing the temperature affect the other variables in the ideal gas law?

According to the ideal gas law, temperature and volume are directly proportional, while pressure and number of moles are inversely proportional. This means that increasing the temperature will increase the volume and decrease the pressure and number of moles.

5. What are some real-life applications of the ideal gas law?

The ideal gas law has various applications in fields such as chemistry, physics, and engineering. It is used to calculate the volume of gas in a container, predict the behavior of gases in different conditions, and design gas systems like boilers and refrigerators.

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