Question

A helium-filled balloon escapes a child's hand at sea level and 20.0∘C . Part A When it reaches an altitude of 3600 m , where the temperature is 5.0 ∘C and the pressure only 0.68 atm , how will its volume compare to that at sea level?

Answer 1

Volume of the balloon would be approximately
`1.4` times the value at sea level, assuming that the content of the balloon behaves like an ideal gas.

Step-by-step explanation:

Assume that the gas in the balloon is an ideal gas. The ideal gas equation would hold:

`P\, V = n\, R\, T`,

Where:

• 
  `P` is the pressure of the gas,
• 
  `V` is the volume of the gas,
• 
  `n` is the quantity of particles in the gas,
• 
  `R` is the ideal gas constant, and
• 
  `T` is the temperature of the gas, as measured on an absolute scale such as the Kelvin scale.

This question implies that the quantity
`n` of gas particles in this balloon stays the same. Let
`V_(1)`,
`P_(1)`, and
`T_(1)` denote the volume, pressure, and temperature at sea level, and let
`V_(2)`,
`P_(2)`, and
`T_(2)` denote the new values of these quantities.

The goal is to find the ratio
`(V_(2) / V_(1))` representing the new volume of the balloon relative to the initial volume. Apply the following steps to find this ratio:

• Rearrange the ideal gas equation to find the relationship between
  `V_(1)`,
  `V_(2)`,
  `P_(1)`,
  `P_(2)`,
  `T_(1)`, and
  `T_(2)`.
• Rearrange the relation from the previous step to find an expression for the ratio
  `(V_(2) / V_(1))` in terms of
  `P_(1)`,
  `P_(2)`,
  `T_(1)`, and
  `T_(2)`.
• Ensure that temperatures are measured in degrees Kelvins (an absolute scale for temperature.) Substitute the values into the expression from the previous step and evaluate to find the value of
  `(V_(2) / V_(1))`.

At the initial pressure and temperature, the ideal gas equation would be:

`P_(1)\, V_(1) = n\, R\, T_(1)`.

Rearrange to obtain:

`\displaystyle (P_(1)\, V_(1))/(T_(1)) = n\, R`.

Similarly, under the new pressure and temperature, this equation becomes:

`P_(2)\, V_(2) = n\, R\, T_(2)`.

`\displaystyle (P_(2)\, V_(2))/(T_(2)) = n\, R`.

Assuming that the quantity of gas particles in this balloon stays the same. The value of
`n\, R` would stay constant regardless of the value of
`P` and
`V`. Hence:

`\displaystyle (P_(1)\, V_(1))/(T_(1)) = n\, R = (P_(2)\, V_(2))/(T_(2))`.

`\displaystyle (P_(1)\, V_(1))/(T_(1)) = (P_(2)\, V_(2))/(T_(2))`.

Rearrange the equation above to find an expression for the ratio
`(V_(2) / V_(1))`:

`\displaystyle (V_(2))/(V_(1)) = (P_(1)\, T_(2))/(P_(2)\, T_(1))`.

To find the temperature on the Kelvin scale, add (approximately)
`273.15` to the value of temperature on the celsius scale:

`T_(1) \approx (273.15 + 20.0)\; {\rm K} = 293.15\; {\rm K}`.

`T_(2) \approx (273.15 + 5.0)\; {\rm K} = 278.15\; {\rm K}`.

The pressure at sea level is
`1\; \text{atm}`. Thus:
`P_(1) = 1\; \text{atm}`. Given that
`P_(2) = 0.68\; \text{atm}`:

`\begin{aligned} (V_(2))/(V_(1)) &= (P_(1)\, T_(2))/(P_(2)\, T_(1)) \\ &\approx \frac{(1\; \text{atm})\, (278.15\; {\rm K})}{(0.68\; \text{atm})\, (293.15\; {\rm K})} \\ &\approx 1.4 \end{aligned}`.

In other words, the new volume would be approximately
`1.4` times that at the sea level.