How many moles of gas must be forced into a 3.1L ball to give it a gauge pressure of 9.8psi at 22degrees C? The gauge pressure is relative to atmospheric pressure. Assume that atmospheric pressure is 14.5psi so that the total pressure in the ball is 24.3psi .
Convert 24.3 psi to atm and use PV = nRT, then solve for n.
24.3 psi is approximately 1.65 atm. Remember T must be in kelvin.
To find the number of moles of gas that must be forced into the ball, we can use the ideal gas law equation:
PV = nRT
Where:
P is the pressure of the gas in atmospheres (atm)
V is the volume of the gas in liters (L)
n is the number of moles of gas
R is the ideal gas constant (0.0821 L·atm/(mol·K))
T is the temperature of the gas in Kelvin (K)
First, let's convert the given temperatures from Celsius to Kelvin:
22 degrees C + 273.15 = 295.15 K
Next, we convert the pressure to atm:
9.8 psi + 14.5 psi = 24.3 psi
24.3 psi * (1 atm / 14.7 psi) = 1.65 atm
Now, we can substitute the values into the ideal gas law equation and solve for the number of moles (n):
(1.65 atm) * (3.1 L) = n * (0.0821 L·atm/(mol·K)) * (295.15 K)
(5.115 atm·L) = (24.23 n)
n = (5.115 atm·L) / (24.23)
n = 0.211 moles
Therefore, approximately 0.211 moles of gas must be forced into the 3.1L ball to give it a gauge pressure of 9.8psi at 22 degrees C.