Pressure on an object as it rises up into the atmosphere

[wiggler115]

how would you find the pressure on an object as it rises up into the atmosphere

ex.
A child's helium-filled balloon escapes at sea level and 20.0°C. It reaches an altitude of 3330 m, where the temperature is 5.0°C and the pressure only 0.65 atm. What is the ratio of its volume at this altitude to its volume at sea level?

 

[mgb_phys]

double post - ignore

 

[astrokreng]

The pressure on an object as it rises up into the atmosphere can be calculated using the ideal gas law, which states that pressure is directly proportional to the number of gas molecules, temperature, and volume. As the object rises in altitude, the number of gas molecules remains constant, but the temperature and volume change.

To find the pressure on an object as it rises up into the atmosphere, we can use the equation P1V1/T1 = P2V2/T2, where P1 is the initial pressure, V1 is the initial volume, T1 is the initial temperature, P2 is the final pressure, V2 is the final volume, and T2 is the final temperature.

In this scenario, P1 is the pressure at sea level (1 atm), V1 is the volume at sea level (the volume of the balloon), T1 is the temperature at sea level (20.0°C + 273.15 = 293.15 K), P2 is the pressure at an altitude of 3330 m (0.65 atm), V2 is the volume at this altitude (unknown), and T2 is the temperature at this altitude (5.0°C + 273.15 = 278.15 K).

Using the given values and rearranging the equation, we can solve for V2: V2 = (P1V1T2)/(P2T1) = (1 atm)(V1)(278.15 K)/(0.65 atm)(293.15 K) = 1.07V1

Therefore, the volume of the balloon at an altitude of 3330 m is 1.07 times the volume of the balloon at sea level. This means that the balloon will expand as it rises into the atmosphere due to the decrease in pressure. The ratio of its volume at this altitude to its volume at sea level is 1.07:1.