How many moles of gas must be forced into a 3.5 L ball to give it a gauge pressure of 9.4 psi at 25 degrees C? The gauge pressure is relative to atmospheric pressure. Assume that atmospheric pressure is 14.7 psi so that the total pressure in the ball is 24.1 psi

Ok so I'm doing PV=nRT solving for n. PV/RT

So I'm converting psi to atm but I'm confused what is all the extra info given. Isn't it 9.4 psi / 14.7 psi x 1 atm

No, if the gauge pressure is 9.4 then the pressure in the ball will be 14.7+9.4 psi = 24.1 psi and that's what the other information tells you. So

24.1 psi x (1 atm/14.7 psi) = ?atm
Then n - PV/RT

Ahhhhh thank you!!!!! I was so frustrated

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Yes, this technique worked. I have how many moles of gas must be forced into a 4.0L ball to give it a pressure of 8.6 psi at 28C. Assume the atm is 14.8 psi so that the total pressure in the ball is 23.4psi

.0821*301.1/4*1.25=
6.36/24.74=.257

To solve this problem, you are correct that you need to convert the pressure from psi to atm. The conversion factor is 1 atm = 14.7 psi, so you can convert the gauge pressure of 9.4 psi to atm using the following calculation:

9.4 psi / 14.7 psi/atm = 0.638 atm

Now, you can use the ideal gas law equation, PV = nRT, to find the number of moles of gas (n) that must be forced into the ball. The formula for the ideal gas law is:

PV = nRT

Where:
P = pressure in atm
V = volume in liters
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin

Given:
P = 0.638 atm (this is the pressure you calculated in atm)
V = 3.5 L (the volume of the ball)
R = 0.0821 L·atm/mol·K (ideal gas constant)
T = 25°C = 25 + 273.15 = 298.15 K (temperature converted to Kelvin)

You can now plug in these values into the ideal gas law equation:

0.638 atm * 3.5 L = n * 0.0821 L·atm/mol·K * 298.15 K

Simplifying the equation:

2.233 L·atm = 24.408215 n

Now, solve for n:

n = (2.233 L·atm) / (24.408215)

n ≈ 0.0914 moles of gas

Therefore, you would need to force approximately 0.0914 moles of gas into the 3.5 L ball to achieve a gauge pressure of 9.4 psi at 25 degrees Celsius.