Ahelium-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

762.2 m³