A helium-filled balloon is launched when the temperature at ground level is 27.8°c and the barometer reads 752 mmhg. if the balloon's volume at launch is 9.47 * 10^{4} 4 l, what is the volume in liters at a height of 36 km, where the pressure is 73.0 mm hg and temperature is 235.0 k? (enter your answer using either standard or scientific notation. for scientific notation, 6.02 x 10^{23} 23 is written as 6.02e23.

The helium may be treated as an ideal gas, so that

(p*V) / T = constant

where

p = pressure

V = volume

T = temperature.

Note that

7.5006 x 10⁻³ mm Hg = 1 Pa

1 L = 10⁻³ m³

Given:

At ground level,

p₁ = 752 mm Hg

= (752 mm Hg) / (7.5006 x 10⁻³ mm Hg/Pa)

= 1.0026 x 10⁵ Pa

V₁ = 9.47 x 10⁴ L = (9.47 x 10⁴ L) * (10⁻³ m³/L)

= 94.7 m³

T₁ = 27.8 °C = 27.8 + 273 K

= 300.8 K

At 36 km height,

P₂ = 73 mm Hg = 73/7.5006 x 10⁻³ Pa

= 9.7326 x 10³ Pa

T₂ = 235 K

If the volume at 36 km height is V₂, then

V₂ = (T₂/p₂) * (p₁/T₁) * V₁

= (235/9.7326 x 10³) * (1.0026 x 10⁵/300.8) * 94.7

= 762.15 m³

Answer: 762.2 m³