A helium-filled balloon is launched when the temperature at ground level is 27.8°c and the barometer reads 752 mmhg. if the ball
oon'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
It’ll be 200 meters West because 500-300=200 and The west direction has the greatest force you could say ... Anytime you have a question asking about displacement you subtract then use the direction with the greatest number ..
