Question

A ship is sailing in the pacific ocean, where the temperature is ${23.4^0}C$ , a balloon has volume $2L$ . What is the volume of the balloon when the ship is in the Indian ocean and the temperature is ${26.1^0}C$.

Answer 1

We know about three fundamental gas laws that define the relationship between pressure, volume, and temperature. These laws are called Boyle’s law, Charles’ Law, and Avogadro’s Law. Charles’ Law states that the volume of a gas increases as temperature increases.

Complete step-by-step solution: We know about three fundamental gas laws that define the relationship between pressure, volume, and temperature. This relationship is given by the equation
$PV = nRT$……$\left( 1 \right)$
where $P =$ Pressure
$V =$ Volume
$n =$ Amount of the substance
$R =$ Ideal gas constant
$T =$ Temperature
From the equation, $\left( 1 \right)$ we can see the relationship between volume and temperature. This relationship is defined by Charles’ law. It states that the volume of a gas increases as temperature increases. In Charles’ law, it is assumed that the pressure and the amount of gas remain constant. Thus, the law can be written as
$\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{T_1}}}{{{T_2}}}$ ……$\left( 2 \right)$
${V_1},{V_2} =$ The volume of the gas
${T_{1,}}{T_2} =$ Temperatures
From the question given. We can substitute the values. We get
${V_1} = 2L$ , ${T_1} = 273 + 23.4 = 296.4K$ and ${T_2} = 273 + 26.1 = 299.1K$
Substituting these values in the equation $\left( 2 \right)$
\eqalign{ & \Rightarrow \dfrac{2}{{{V_2}}} = \dfrac{{296.4}}{{299.1}} \cr & \Rightarrow {V_2} = \dfrac{{2 \times 299.1}}{{296.4}} \cr}
Solving it, we get the value of ${V_2} = 2.018L$

Hence the volume of the balloon when the ship is in Indian Ocean is 2.018L.

Note: It is important to note that here we have changed the units of temperature. Temperatures, in the question, were given in degrees Celsius. We have converted the given temperature into Kelvin by using the following conversion
$0^0C + {\text{ }}273.15{\text{ }} = K$