How Much Air Escapes from a Heated Compressed Air Cylinder?

[risingabove]

Homework Statement

A cylinder containing 19 kg of compressed air at a pressure 9.5 times that of the atmosphere is kept in a store at 7 degrees Celsius. When it is moved to a workshop where the temperature is 27 degrees Celsius a safety valve on the cylinder operates, releasing some of the air. If the valve allows air to escape when its pressure exceeds 10 times that of the atmosphere, Calculate the mass of air that escapes..

Homework Equations

PV= nRT

where P - Pressure
V- Volume
n- number of moles
R- universal gas constant (8.314 J / mol. K )
T - temperature on Kelvin

Moles = mass / Mr

pressure1 9.5 x 760 mmHg = 7220 mmHg

Temperature1 7°C = 280 K

pressure2 10 x 760 mmHg = 7660 mmHg

Temperature 27°C = 300 K

The Attempt at a Solution

I assumed that the volume was constant and at STP 22.4 x 10^-3 m³

rearranging the equation PV = nRT, placing n as the subject

With Pressure 1 and Temperature 1

n = 7220(22.4 x 10^-3) / 8.314(280)

n = 0.0694 moles

With Pressure 2 and Temperature 2

n = 7660(22.4 x 10^-3) / 8.314(300)

n = 0.0687 moles

Rearranging the equation for moles to find Mr, using the initial mass and moles from pressure 1 and temperature 1

Mr = 19 kg / 0.0694 moles

Mr = 274

using the Mr value calculated to find mass from pressure 2 and temperature 2

274 x 0.0687 = 18.82 kg

subtracting the above value from the initial mass 19

19 - 18.82= 0.18 kgIm not sure if i worked this out right...or if it makes any sense at all but it was as close i could of gotten in attempting this problem... please any assistance of guidance will be grateful...

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Simon Bridge]

Not bad - modelling the air as an ideal gas ... the ideal gas equation takes form:
I think you did a lot more work than you needed to. It helps t do all the algebra before you plug the numbers in.

The trick with this sort of problem is to realize that you can divide the ideal gas equations for each state.

Step 1. lynchpin: write out PV=nRT in terms of the total mass of gas present M: M=nm: m=molar mass of the gas.

Thus: $PV=(R/m)MT$

This is useful because you have the amount of gas as a total mass, and writing it out this way avoids having to convert moles a couple of times. Since R/m is a constant, it will cancel out in the next step so we don't even need to know the values of R and m.

step 2. if state 2 is the final state, state 1 is the initial state: divide the ideal gas equations out like this: $$\frac{P_2V_2=(R/m)M_2T_2}{P_1V_1=(R/m)M_1T_1}$$... you should be able to simplify the equation, remember that $V_2=V_1$ and $[P]_{atmos}\propto [P]_{mmHg}$ ... and you are not actually looking for $M_2$ yu are looking for the mass lost.

 

[risingabove]

ooo my... that's embarrassing...i always end up doing extra work...or over thinking it...thanks soo much, followed your guidance and got M2 to be 18.67kg... by subtraction from 19 .. Mass escape worked out to be 0.33kg

 

[Simon Bridge]

Many students feel more comfortable with numbers but the algebra is your friend.
It's too easy to make rounding off errors or lose track if you don't do the algebra first.
That's the main lesson here.

 

[HalfLostAndSearching]

I appreciate your attempt at solving this problem. However, I believe there are a few errors in your calculations. First, the volume of the cylinder is not given, so we cannot assume it is constant. Additionally, the ideal gas law (PV = nRT) cannot be used in this scenario because the gas is not at constant temperature and pressure.

To solve this problem, we can use the combined gas law, which states that PV/T = constant. We can also assume that the number of moles of gas remains constant. Therefore, we can set up the following equation:

(P1V1)/T1 = (P2V2)/T2

We can rearrange this equation to solve for V2:

V2 = (P1V1T2)/(T1P2)

Now, let's plug in the given values:

V2 = (7220 mmHg)(22.4 x 10^-3 m^3)(300 K) / (280 K)(7660 mmHg)

V2 = 0.0201 m^3 or 20.1 L

This is the volume of the cylinder after the air has escaped. To find the mass of the air that escaped, we can use the ideal gas law:

PV = nRT

Solving for n (number of moles):

n = (PV)/(RT)

n = (7660 mmHg)(0.0201 m^3) / (8.314 J/mol K)(300 K)

n = 0.00982 moles

Now, we can use the given mass and number of moles to find the molar mass of the gas:

Mr = mass/n

Mr = (19 kg)/(0.00982 moles)

Mr = 1936 g/mol

Finally, we can use this molar mass to find the mass of the air that escaped:

Mass of air that escaped = (0.00982 moles)(1936 g/mol)

Mass of air that escaped = 19.01 kg

Therefore, the mass of air that escaped from the cylinder is approximately 19.01 kg. I hope this helps clarify the solution to this problem. Keep up the good work in your scientific studies!