Question

A balloon is inflated with 2.42 L of helium at a temperature of 27 degrees celsius. Later, the volume of the balloon has changed to 2.37 L at a temperature of 19 degrees celsius and a pressure of 99.7 kPa. What was the pressure when the balloon was inflated?

Answer 1

Answer:

$P_1=100.3kPa$

Explanation:

Hello,

In this case, we use the combined law of gases which helps us to understand the changing pressure-volume-temperature behavior as shown below:

$\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}$

In this case, we are asked to compute the initial pressure before the ballon inflation, considering we must use absolute temperature units, therefore:

$P_1=\frac{P_2V_2T_1}{T_2V_1} =\frac{99.7kPa*2.37L*(27+273)K}{(19+273)K*2.42L} \\\\P_1=100.3kPa$

Best regards.

Answer 2

Answer:

Pressure when the balloon was inflated is 100kPa

Explanation:

It is possible to find the pressure of a gas in a state using combined gas law:

P₁V₁/T₁ = P₂V₂/T₂

Where P is pressure, V is volume and T is absolute temperature (°C+273.15) of 1, initial state, and 2, final state.

Replacing:

P₁2.42L/(27°C+273.15) = 99.7kPa*2.37L/(19°C+273.15)

P₁ = 100kPa

Pressure when the balloon was inflated is 100kPa