Question

A helium weather balloon is released from ground level where the temperature is 20⁰ C. The balloon rises to a maximum height of 8.0 miles above the ground. At this height, the balloon expands to a volume that is 20 times greater than at the release point, and the pressure of the helium gas is 25 times smaller than at the release point. What is the final temperature, in ⁰F, of the gas at the maximum height of the balloon?

Answer #1

We know the Ideal Gas law states that

Where

P | Pressure |

V | Volume |

T | Temperature |

n | No of Moles |

R | Ideal Gas Constant |

Now for a fixed amount of gas, i.e. n = constant; we infer that

So we can write:

...... (Eqn 1)

Now from the given data we have,

**T _{1} = 20°C = 293.15 K** (Subscript 1
means at ground, Subscript 2 means at final height)

**P _{2} = P_{1}/25 and V_{2} =
20V_{1}**

Substituting into Eqn 1 we get

Therefore;

In Celcius **T _{2} = -38.63°C**

The Conversion Formula is: **T _{(°F)} =
T_{(°C)} × 9/5 + 32**

Hence temperature at 8miles height is **T _{2} =
-38.63° × 9/5 + 32 = -37.53°F** (ANSWER)

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