Question

Larisa pumps up a soccer ball until it has a gauge pressure of 61 kilopascals. The volume of the ball is 5.2 liters. The air temperature is 32°C, and the outside air is at standard pressure. How many moles of air are in the ball?

Answer 1

The total ball pressure is sum of gauge pressure and atmospheric pressure

We know that atmospheric pressure = 101.325 kPa

so the total ball pressure = 61 kPa + 101.325 kPa = 162.325 kPa

Now we will use ideal gas equation as

PV = nRT

P = pressure = 162.325 kPa

V = 5.2 L

R = gas constant = 8.314 kPa  L / mol K

T = 32 C = 273.15 + 32 = 305.15 K

n = moles = ?

Moles = PV / RT = 162.325 kPa X 5.2 / 305.15 X 8.314 = 0.332 moles

Answer 2

A 5.2 L-soccer ball with a gauge pressure of 61 kPa at 32 °C contains 332 moles of air.

Larisa pumps up a soccer ball until it has a gauge pressure of 61 kilopascals. The total pressure (P) inside the ball is the sum of the gauge pressure and the atmospheric pressure.

$P = P_g + P_{atm} = 61 kPa + 101.325 kPa = 162 kPa$

The air temperature (T) is 32 °C. We will convert this value to Kelvin using the following expression.

$K = \° C + 273.15 = 32\° C + 273.15 = 305 K$

The volume (V) of the ball is 5.2 L. We can calculate the moles (n) of air inside the ball using the ideal gas equation.

$P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{(162 \times 10^{3}Pa ) \times 5.2L}{\frac{8.134J}{mol.K} \times 305K} = 332 mol$

A 5.2 L-soccer ball with a gauge pressure of 61 kPa at 32 °C contains 332 moles of air.

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