A particular sample of neon gas has a pressure of 13.2 atm and volume of 2.0 L. What will the new pressure be of the sample of neon if the container's volume is expanded to 3.0 L?

Answer 1

Since no temperature is mentioned, it is assumed to be constant, and so we are dealing with Charles' Law. This can be written as P1V1 = P2V2

P1 = 13.2 atm

V1 = 2.0 L

P2 = ?

V2 = 3.0 L

P1V1 = P2V2

(13.2 atm)(2.0 L) = (P2)(3.0 L)

P2 = 13.2 x 2.0 / 3.0

P2 = 8.8 atm

Answer 2

Quoted below is a question about how volume affects pressure under the conditions of constant temperature and constant molecule count. The fact that it's Neon molecules involved is not pertinent, it could be any type of gaseous molecule as it is a gas law. The initial conditions say, "A particular sample of Neon gas has a pressure of 13.2 atmospheres and a volume of 2 L, what will the new pressure be of the sample of Neon if the container volume is expanded 3.0 L?"

Background information: Pressure is a continuous force exerted against a unit area of a container's walls.

1 atmosphere is a standard unit of pressure, which is approximately equal to the atmospheric pressure at sea level.

If you have a pressure in a given volume under old conditions at a given temperature and the same amount of gas molecules, (P1)(V1), as long as the temperature and the amount of gas remains the same under a new condition, if you change the volume under a new condition, the new pressure must change as well.

So, if the initial pressure is 13.2 atm in a volume of 2L (High pressure in a small container), What would happen to the pressure if you increased the volume to 3L? (Common sense says, the pressure would _________?).

This concept is visually like a giant syringe working with gaseous molecules not liquids. initially, the syringe had a smaller volume of a given gas in it, but when the plunger was pulled back, and the volume increased, as long as the temperature or number of gaseous particles didn't change, under the new condition of a bigger volume, the pressure of the same amount of molecules banging against wall walls would be less.

Mathematically:

(P1)(V1)= (P2)(V2) at constant temperature and the same number of gas molecules. Plug in the numbers,

and solve for P2:

(P1)(V1)/V2 = P2

(13.2 atm)(2L)/(3L) = P2: (13.2 atm)(2L)/(3L) = P2, since the liters cancel, the answer has the units of atms

(13.2 atm)(2L)/(3L) = P2 , or 8.8 atm = P2

Answer 3

This is a problem involving Boyle's Law, which relates the pressure and the volume by the equation

PV = k, where k is a constant.

This means for initial conditions P1 and V1, P1*V1 = k, and for final conditions P2 and V2,

P2*V2 = k

Since, for a given temperature and amount of gas, the constant k is the same for both, then the two can be equated to give

P1*V1 = P2*V2. You are given that P1 = 13.2 atm and V1 = 2.0 L.

You are also given V2 = 3.0 L. Plug in the numbers and solve for P2 to get

13.2 atm * 2.0 L = P2 * 3.0 L. P2 = (13.2atm *2.0 L)/(3.0 L) = 8.8 atm