Shaw W.
asked • 05/01/20hot air balloon is filled with 1.19 × 10 to the 6th of an ideal gas on a cool morning (11 °C). The air is heated to 129 °C. What is the volume of the air
Bishal N. answered • 05/01/20
Experienced High School Math, Physics, and Chemistry Tutor.
According to Charles law V/T = constant
Let V1 be the volume of air in balloon at temperature T1 and V2 be the volume of the air in balloon after heating to temperature T2
T1 = 110 C = (11+273)K = 284 K
T2= 1290C = (129+273) = 402K
V1/T1 = V2/T2
V2 =( V1 *T2)/T1
= (1.19*106*402)/284
= 1.68*106 L
Jesse E. answered • 05/01/20
Masters in Chemistry with 4+ years of teaching experience.
We will use Charles Law in the following form:
V1/T1=V2T2
I am going to assume the volume is in Liter only to have units but in this case, the actual units do not matter.
1.0 x 106 L / (11 + 273) = V2/(129 + 273)
Rearrange the equation.
V2 = (1.0 x 106 L / 11 + 273) ) x (129 + 273) = 1.4 x 106 L.
Jesse E.
Thanks for that. I completely spaced out about that detail.05/01/20
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