How many moles of gas must be forced into a 3.9 ball to give it a gauge pressure of 9.8 at 25? The gauge pressure is relative to atmospheric pressure. Assume that atmospheric pressure is 14.5 so that the total pressure in the ball is 24.3 .
Express your answer using two significant figures.
How many moles of gas must be forced into a 3.9 ball to give it a gauge pressure of 9.8 at 25? The gauge pressure is relative to atmospheric pressure. Assume that atmospheric pressure is 14.5 so that the total pressure in the ball is 24.3 .
Express your answer using two significant figures
To solve this problem, we need to use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atm)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
First, let's convert the temperature from degrees Celsius to Kelvin:
T(K) = T(°C) + 273.15
T(K) = 25 + 273.15
T(K) = 298.15 K
Now, we can rearrange the ideal gas law equation to solve for the number of moles (n):
n = PV / RT
Since the problem provides the gauge pressure, we need to convert it to absolute pressure by adding the atmospheric pressure:
P(total) = P(gauge) + P(atmospheric) = 24.3 + 14.5 = 38.8 atm
Now we can substitute the values into the equation:
n = (P(total) * V) / (R * T)
Substituting the given values:
n = (38.8 * 3.9) / (0.0821 * 298.15)
Calculating the value:
n ≈ 6.52
Therefore, approximately 6.52 moles of gas must be forced into the 3.9 L ball to give it a gauge pressure of 9.8 atm at 25°C.