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A balloon containing 2.00 mols of an ideal gas has a volume of 0.0240 m3 (1 m3 = 1000L) at a temperature of 149 K. What is the pressure of the gas in the balloon in kPa? Report your answer to the correct number of significant figures. R = 8.3145 L kPa / mol K

User Enloz
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Final answer:

The pressure of a balloon containing 2.00 mols of an ideal gas at a volume of 0.0240 m3 and a temperature of 149 K can be calculated using the ideal gas law with a pressure value of 5.129 kPa.

Step-by-step explanation:

The question is about calculating the pressure of an ideal gas in a balloon using the ideal gas law equation. Since we have been given the number of moles of the gas, the volume of the balloon, and the temperature, we can apply the ideal gas law, PV = nRT, where P is pressure, V is volume, n is number of moles, R is the ideal gas constant, and T is temperature in Kelvin.

Using R = 8.3145 L kPa / mol K, and converting the volume to liters, we can solve for the pressure in kPa.

Example Calculation:

Number of moles (n) = 2.00 mols

Volume (V) = 0.0240 m3

= 24.0 L (since 1 m3

= 1000 L)

Temperature (T) = 149 K

Ideal gas constant (R) = 8.3145 L kPa / mol K

Substitute the values into the ideal gas law equation:

P = nRT / V

P = (2.00 mols) * (8.3145 L kPa / mol K) * (149 K) / (24.0 L)

P = 5.129 kPa (rounded to three significant figures)

The pressure of the gas in the balloon is, therefore, 5.129 kPa.

User Pekka
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