Assuming the question require we get the new volume at the altitude of 4.05 Km
1028.09 L
We are given;
Initial volume of Helium, V1 = 793 L
Initial temperature, T1 = 16.9 °C (But K = °C + 273.5)
, thus, T1 = 290.05 K
Initial pressure, P1 = 759 mmHg
New temperature at the height of 4.05 km, T2 = -7.1°C + 273.15
= 266.05 K
Pressure at the Height of 4.05 Km, P2 = 537 mmHg
We are required to calculate the volume attained by the balloon while at the height of 4.05 Km
We are going to use the combined gas law.
According to the combined gas law; [tex]\frac{P1V1}{T1} =\frac{P2V2}{T2}[/tex]
Therefore, making V2 the subject of the equation;
[tex]V2 =\frac{P1V1T2}{P2T1}[/tex]
[tex]V2 =\frac{(759mmHg)(793L)(266.05K)}{(537mmHg)(290.05K)}[/tex]
[tex]V2 =1028.09[/tex]
Therefore, the volume of the balloon at the height of 4.05 km will be 1028.09L
