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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.

User Depiction
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Final answer:

The volume of the helium-filled balloon at a height of 36 km, under the conditions given, can be calculated using the Combined Gas Law, and it is approximately 9.47 × 10^7 liters.

Step-by-step explanation:

To calculate the volume of the helium-filled balloon at a height of 36 km, where the pressure is 73.0 mmHg and the temperature is 235.0 K, we can use the Combined Gas Law. This law combines the three gas laws: Boyle's Law, Charles's Law, and Gay-Lussac's Law. It is represented by the formula:



P1 * V1 / T1 = P2 * V2 / T2



Where:

P1 = initial pressure

V1 = initial volume

T1 = initial temperature (in Kelvin)

P2 = final pressure

V2 = final volume (unknown)

T2 = final temperature (in Kelvin)



To apply the Combined Gas Law to our scenario, we must first convert the initial temperature to Kelvin:



T1 = 27.8°C + 273.15 = 300.95 K



Now we can plug in the values:



(752 mmHg * 9.47 × 10^4 L) / 300.95 K = (73.0 mmHg * V2) / 235.0 K



By rearranging the formula to solve for V2, the final volume, we get:



V2 = (752 mmHg * 9.47 × 10^4 L * 235.0 K) / (73.0 mmHg * 300.95 K)



After performing the calculations:



V2 = 9.47 × 10^7 L approximately (the exact answer depends on the precision of the calculations and the rounding)



Thus, the volume of the balloon at 36 km would be approximately 9.47 × 10^7 liters.

User Akhneyzar
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1 vote
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
User Volod
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