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# On a ship sailing in the Pacific ocean where temperature is 23.4, is filled with 2L air. What will be the volume of the balloon when the ship reaches Indian ocean, Where temperature is 26.1?

Last updated date: 20th Jun 2024
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Hint:The physical principle known as Charles' law states that the volume of a gas equals a constant value multiplied by its temperature as measured on the Kelvin scale (zero Kelvin corresponds to -273.15 degrees Celsius).

Complete step-by-step solution:Charles' Law is an experimental gas law that describes how gases tend to expand when heated. The law states that if a quantity of gas is held at a constant pressure, there is a direct relationship between its volume and the temperature, as measured in degrees Kelvin.
The atmospheric pressure on the gas in both the oceans will be constant .
Let the volume and temperature in pacific ocean be represented by ${V_p}$ and ${T_p}$. And let the volume and temperature in Indian ocean be represented by ${V_I}$ and ${T_I}$.
So, the given data of the question can be filtered out as:
${V_p} = 2L$ and ${T_p} = 273 + 23.4 = 296.4K$
${V_I} = ?$ and ${T_I} = 273 + 26.1 = 299.1K$
According to Charles law: $\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{T_1}}}{{{T_2}}}$
So, we have $\dfrac{{{V_p}}}{{{V_I}}} = \dfrac{{{T_p}}}{{{T_I}}}$
Substituting the values of volume and temperature of pacific and Indian ocean from above in this equation:
${V_I} = \dfrac{{{V_P}{T_I}}}{{{T_P}}}$
$\Rightarrow {V_I} = \dfrac{{2L \times 299.1K}}{{296.4K}}$
$\Rightarrow {V_I} = 2L \times 1.009L = 2.018L$
$\therefore {V_I} = 2.018L$

Hence the volume of the balloon when the ship reaches Indian ocean, Where temperature is 26.1 will be 2.018L.

Note:Remember that the example of Charles law in real life is tyres of untouched vehicles get deflated during freezing winter days while get inflated in hot summer days. This unusual behaviour is because of Charles's law. In winter due to low temperatures, the air inside a tyre gets cooler, and they shrink. While on hot days,the air expands with temperature.