A 15.0L sample of argon has a pressure of 28.0 atm . What volume would this gas occupy at 9.90 atm? Assume ideal behavior.
Isn't this just like the last post?
P1V1 = P2V2
Yes I am so sorry
no
A sample of argon gas at a pressure of 0.756 atm and a temperature of 25.5°C, occupies a volume of 13.2 liters. If the gas is compressed at constant temperature to a volume of 3.33 liters, the pressure of the gas sample will be
atm.
To find the volume that the gas would occupy at a different pressure, you can use the ideal gas law equation: PV = nRT.
Here, P represents the initial pressure, V represents the initial volume, n represents the number of moles of gas, R is the ideal gas constant (0.0821 L·atm/(mol·K)), and T is the temperature in Kelvin.
Since we are assuming ideal behavior, we can assume that n and T remain constant.
Let's assign the following values to the variables:
P1 = 28.0 atm (initial pressure)
V1 = 15.0 L (initial volume)
P2 = 9.90 atm (new pressure)
V2 = ? (new volume)
We can set up the equation like this:
(P1 * V1) = (P2 * V2)
Plugging in the values we know:
(28.0 atm * 15.0 L) = (9.90 atm * V2)
Now, solve for V2:
V2 = (28.0 atm * 15.0 L) / 9.90 atm
V2 = 420 L / 9.90
V2 ≈ 42.42 L
Therefore, the volume that the gas would occupy at 9.90 atm is approximately 42.42 L.