Answer:
[tex]P_1=100.3kPa[/tex]
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:
[tex]\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}[/tex]
In this case, we are asked to compute the initial pressure before the ballon inflation, considering we must use absolute temperature units, therefore:
[tex]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[/tex]
Best regards.
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
