i have two questions.

1. generally speaking, if the pressure of a gas is tripled, what will happen to its volume?

2. If I take 2000L of gas at STP and squish it into a gas 15.0L gas cylinder, what will be the pressure inside the cylinder, assuming the temperature is kept constant?

(1) assuming the gas is ideal, the volume becomes 1/3 of the original.

(2) recall that STP, pressure is at 1 atm. assuming the gas is ideal,
P1*V1 = P2*V2
where
P1 = initial pressure
V1 = initial volume
P2 = final pressure
V2 = final volume
substituting,
1*2000 = (P2)*15
P2 = 2000/15
P2 = 133.33 atm

hope this helps~ :)

thank you very much!

if you could help me with this as well i would be very appreciative!

A steel container holds 750.L of O2 gas at STP.
a) how many moles of O2 does the container hold?
b)how many grams of O2 does the container hold?
c) if the pressure were reduced by removval of gas until it was 380.mm Hg, what mass of gas would remain in the tank?

1. To understand the relationship between pressure and volume of a gas, we can refer to Boyle's Law, which states that at a constant temperature, the product of pressure and volume of a gas is constant. Mathematically, it can be expressed as P1 * V1 = P2 * V2, where P1 and V1 are the initial pressure and volume of the gas, and P2 and V2 are the final pressure and volume.

In this case, if the pressure of a gas is tripled, we can assume that the initial pressure (P1) becomes three times its original value, and we need to find how the volume (V2) changes. Using Boyle's Law, we can rearrange the equation as V2 = (P1 * V1) / P2. Plugging in the respective values, we get V2 = (3 * V1) / 1. Therefore, when the pressure is tripled, the volume of the gas will decrease to one-third of its original volume.

2. To determine the pressure inside the cylinder, we can use the ideal gas law, which states that the product of pressure (P), volume (V), and temperature (T) of a gas is directly proportional to the number of moles (n) and the ideal gas constant (R). The equation can be written as PV = nRT.

Given that the temperature is kept constant and the gas is being squished from 2000L to 15.0L, the volume (V2) becomes 15.0L. Let's assume that the number of moles (n), remains constant. Therefore, the equation can be simplified as P1 * V1 = P2 * V2.

Plugging in the values, we have P1 * 2000L = P2 * 15.0L. Rearranging the equation, we find that P2 = (P1 * 2000L) / 15.0L. Since we know the initial pressure (P1) is at standard temperature and pressure (STP), which is 1 atmosphere (atm), we can substitute P1 = 1 atm.

Now, we can calculate P2 = (1 atm * 2000L) / 15.0L. Simplifying the equation, P2 ≈ 133.33 atm. Therefore, the pressure inside the cylinder would be approximately 133.33 atmospheres if the temperature is kept constant while compressing the gas.