Question

A balloon filled with helium gas occupies 2.50 L at 25 degrees celcius and 1.00 atm. When released, it rises to an atlitude where the temperature is 18 degrees celcius and the pressure is only 0.80 atm. Calculate the new volume of the balloon.

Answer 1

This task can be solved by using Ideal gas law. The ideal gas law is the equation of state of a hypothetical ideal gas. It is a good approximation to the behaviour of many gases under various conditions, although it has several limitations. The ideal gas law is often introduced in its common form:

$$PV = nRT$$

where $P$ is the pressure of the gas, $V$ is the volume of the gas, $n$ is the amount of substance of gas (also known as number of moles), $T$ is the temperature of the gas and $R$ is the ideal, or universal, gas constant, equal to the product of the Boltzmann constant and the Avogadro constant.

In SI units, $P$ is measured in pascals, $V$ is measured in cubic metres, $n$ is measured in moles, and $T$ in kelvin (273.15 Kelvin = 0.00 degrees Celsius). $R$ has the value $8.314\mathrm{J}\cdot\mathrm{K}^{-1}\cdot\mathrm{mol}^{-1}$ or $0.08206\mathrm{L}\cdot\mathrm{atm}\cdot\mathrm{mol}^{-1}\cdot\mathrm{K}^{-1}$ if using pressure in standard atmospheres (atm) instead of pascals, and volume in liters instead of cubic metres.

In this task amount of helium is constant, R is constant too. Volume, temperature and pressure is changeable, but two last one is given:

For both cases:

$$\mathrm{PV} = \mathrm{nRT}$$ $$\mathrm{nR} = \mathrm{PV}/\mathrm{T} \text{ (nR is const)}$$ $$\mathrm{So} \quad \mathrm{P_1V_1/T_1} = \mathrm{P_2V_2/T_2}$$ $$\mathrm{V_2} \text{ can be found as: } \mathrm{V_2} = \mathrm{P_1V_1T_2/T_1P_2}.$$ $$\mathrm{V_2} = 1.00 * 2.50 * (273 + 25) / (273 + 18) * 0.80 = 3.2 \mathrm{L}$$