# AP Question involving Ideal Gas Law - Thermal Physics - Pressure

• Mafi0z
In summary, the conversation involved solving a 1996 AP Physics B Free Response Question regarding the Ideal Gas Law, Thermal Physics, and Pressure. The problem involved a cylindrical can filled with an ideal gas that was heated and then cooled, with a loose cap that allowed some gas to escape before being tightened. The conversation included attempts at solving parts a, b, and c of the problem, with varying methods and equations being used. The correct solution for part c was found to be 0.00542 moles of air entering the can.
Mafi0z
[SOLVED]AP Question involving Ideal Gas Law - Thermal Physics - Pressure

I am having some trouble with an old ap problem(PART C ONLY), see below:

## Homework Statement

1996 AP PHYSICS B Free Response Question:
The inside of the cylindrical can has a cross sectional area of 0.005 m^3 and length of 0.15 m. The can is filled with an ideal gas and covered with a loose cap. The gas is heated to 363 K and some is allowed to escape from the can so that the remaining gas reaches atmospheric pressure (1 x 10^5 Pa). The cap is now tightened, and the gas is cooled to 298 K.

a. What is the pressure of the cooled gas?
b. Determine the upward force exerted on the cap by the cooled gas inside the can.
c. If the cap develops a leak, how many moles of air would enter the can as it reaches a final equilibrium at 298 K and atmospheric pressure? (assume that air is an ideal gas)

P1/T1=P2/T2
P=F/A
PV=nRT

## The Attempt at a Solution

a) i figured this one out..
P1/T1=P2/T2
P2 = ~82093 Pascals

b) got this one too..
P = F/A
F = PA
F = ~410.5 Newtons

c) cannot get this one..
the volume of the gas that will enter into the leak is equal to the volume of the gas that escaped when the can was heated, correct? someone has said that this part will involve the volume of an ideal gas at STP, which is 22.4 Liters.
I am not sure what to do..

Attempt at part c - THIS IS INCORRECT, SEE BELOW FOR PROPER WORK
V1/T1=V2/T2
(0.005*0.15)/298=V2/363
V2= 9.136 x 10^-4 m^3

V2-V1 = V(escaped)
V(escaped) = 1.92 x 10^-4 m^3

PV(escaped)=nRT
n=PV(escaped)/RT
n= ((1x10^5)(1.92x10^-4)) / ((8.31)(298))
n= 0.0078 moles ??

Any help, explanations, and/or answers will be greatly appreciated.
Thank you!

Last edited:
For c) note that everything is an ideal gas.

You calculated the underpressure inside the can. That would be 82093 Pa.

But that is an underpressure.

Consider then what volume it will occupy at 100,000 Pa. or 1 atm.

Armed with that you know how much air must have entered to equalize the pressure to 100,000.

Knowing the volume of air entered at 298K allows you to determine the number of moles that entered.

hmm not sure i understand.

do you mean:

P1V1=P2V2 with P1 = 82093 and P2 = 1atm?
V2 = ((82093)(0.005*0.15)) / (1x10^5)
V2 = 6.157x10^-4 m^3

then do the n = PV/RT with V = V2 = 6.157x10^-4 and P = 1atm and T =298 ?
n= 0.0248 moles

still doesn't seem right.. although i am sure i didn't follow your advice, since i didn't really get it
=[

Mafi0z said:
hmm not sure i understand.

do you mean:

P1V1=P2V2 with P1 = 82093 and P2 = 1atm?
V2 = ((82093)(0.005*0.15)) / (1x10^5)
V2 = 6.157x10^-4 m^3

then do the n = PV/RT with V = V2 = 6.157x10^-4 and P = 1atm and T =298 ?
n= 0.0248 moles

still doesn't seem right.. although i am sure i didn't follow your advice, since i didn't really get it
=[

Not quite.

At the under pressure how much might you have needed to reduce the volume in order to equalize the pressure back to 1 atm. You got that part OK.

(1atm)*V = .82atm*(.005)(.15)
That gives the 6.1*10-4

Subtract that from the overall volume 7.5*10-4 - 6.1*10-4 = 1.4*10-4 = ΔV

This change in volume is what needs to be supplied in outside air then right?

Oh ok i got that part now(ΔV). thanks.
so basically what that's asking is what change in volume will occur when the pressure increases from 82093 to 100,000, right?
I forgot to subtract the overall volume, not sure why.. i did on my first attempt.

Anyway, so now would I still use n = PV/RT ? since n = moles..
T = 298 and P = 1atm?

Note: So temperature doesn't change, since it has already been cooled to 298k, and the equilibrium point is also 298k, correct?

Last edited:
Mafi0z said:
Oh ok i got that part now(ΔV). thanks.
I forgot to subtract the overall volume, not sure why.. i did on my first attempt.

Anyway, so now would i still use n = PV/RT ? since n = moles
T = 298 and P = 1atm?

Yes. That was my approach.

Thanks a lot for the help, LowlyPion! :D

.00542 moles of air enters the can

P1V1=P2V2
V2 = ((82093)(0.005*0.15)) / (1x10^5)
V2 = 6.157x10^-4 m^3

ΔV = V(can) - V2
ΔV = (0.005*0.15) - 6.157*10-4
ΔV = 1.343x10^-4

n=PV/RT = ((100000)(1.343x10^-4)) / ((8.31)(298))
n = 0.00542 molesThere are also other ways of doing the problem.
A user(codygo) in the physics channel on freenode irc pointed some out for me..

For those interested:

Alternative solution:

PV/nRT (initial) = PV/nRT (final)
since T is the same (298) and so is V (0.005*0.15), they can be eliminated
the formula now becomes

Pi/ni=Pf/nf

to find ni use ni=PV/RT = ((82093)(0.005*.15)) / ((8.31)(298)) = 0.02486
(Must use combination of either the cooled values, or the heated values - So 100,000Pa and 363K - OR 82093Pa and 298K -i used the cooled ones above - either combo will work..)nf = Pf*ni/Pi
nf = ((100,000)(0.02486)) / (82093)
nf = 0.030286

Amount of air entered = nf-ni
nf-ni = (0.030286)-(0.02486) = 0.00542 moles
:D

Thanks again,
~mafi0zUPDATE: i just realized the mistake i was making in my first attempt at part c..i was using the wrong temperatures..i had T1 and T2 reversed!
It should have been:

V1/T1=V2/T2
(0.005*0.15)/393=V2/268
V2= 6.157 x 10^-4 m^3

Vi - V2 = V(escaped)
V(escaped) = (0.005*0.15) - (6.157 x 10^-4)
V(escaped) = 1.343x10^-4

PV(escaped)=nRT
n=PV(escaped)/RT
n= ((1x10^5)(1.343x10^-4)) / ((8.31)(298))
n= 0.00542 moles

:D

So, like i said, there are many ways of doing this problem..

Last edited:
And think of what you learned along the way.

Good luck.

## 1. What is the Ideal Gas Law?

The Ideal Gas Law is a mathematical equation that describes the relationship between the pressure, volume, temperature, and amount of a gas in a closed system. It is typically written as PV = nRT, where P is pressure, V is volume, n is the number of moles of gas, R is the gas constant, and T is temperature.

## 2. How is the Ideal Gas Law used in thermal physics?

The Ideal Gas Law is used in thermal physics to understand the behavior of gases at different temperatures and pressures. It can be used to calculate the volume or pressure of a gas at a given temperature, or to determine the change in one of these variables when the others are changed. It is also used to study the relationships between different properties of gases, such as the relationship between temperature and pressure.

## 3. What is the significance of the gas constant (R) in the Ideal Gas Law?

The gas constant (R) is a constant value that is used to relate the macroscopic properties of gases (pressure, volume, temperature) to the microscopic properties (number of particles, their mass, and the way they interact). It allows us to use the Ideal Gas Law to make predictions about the behavior of gases, regardless of the specific gas being studied.

## 4. How does the Ideal Gas Law relate to the kinetic theory of gases?

The Ideal Gas Law is based on the kinetic theory of gases, which states that gases are made up of particles in constant motion. The law uses the average kinetic energy of these particles to relate the temperature of a gas to its pressure and volume. The kinetic theory of gases helps to explain the behavior of gases in terms of the motion and interactions of their particles.

## 5. Can the Ideal Gas Law be applied to real-world situations?

While the Ideal Gas Law is a useful tool for understanding the behavior of gases, it is an idealized equation and does not always accurately predict the behavior of real gases. In real-world situations, factors such as intermolecular forces and the size of the gas particles can affect the behavior of gases. However, the Ideal Gas Law can still be used as a good approximation in many situations, especially at high temperatures and low pressures.

Replies
8
Views
1K
Replies
2
Views
2K
Replies
2
Views
2K
Replies
16
Views
2K
Replies
3
Views
9K
Replies
9
Views
2K
Replies
12
Views
1K
Replies
5
Views
1K
Replies
8
Views
1K
Replies
12
Views
1K