Gas Law Practice Problem: Volume Changes with Temperature and Pressure

How does the volume of a gas change with temperature and pressure?

According to Ideal Gas Law, "The Volume of a given mass of gas is inversely proportional to the applied Pressure and directly proportional to the applied temperature." How does this principle work in a practical scenario?

Answer:

When a helium-filled balloon at sea level has a volume of 2.1L at 0.998 atm and 36°C, and it rises to an elevation where the pressure is 0.900 atm and the temperature is 28°C, the new volume of the balloon will be 2.268 L.

Gas laws, particularly the Ideal Gas Law, provide a way to understand how gases behave under different conditions of temperature and pressure. In this scenario, the volume of the gas in the balloon changes as it rises to a higher elevation and experiences a change in pressure and temperature.

According to the Ideal Gas Law equation: "(P₁ * V₁) / T₁ = (P₂ * V₂) / T₂", where P represents pressure, V represents volume, and T represents temperature. By applying this equation to the given data, we can calculate the new volume of the balloon when it reaches the new elevation.

Initial data: Pressure (P₁) = 0.998 atm Volume (V₁) = 2.1 L Temperature (T₁) = 36°C = 309 K Final data: Pressure (P₂) = 0.900 atm Temperature (T₂) = 28°C = 301 K Volume (V₂) = ?

By rearranging the Ideal Gas Law equation and plugging in the given values, we can calculate the new volume of the balloon as follows: V₂ = (P₁ * V₁ * T₂) / (T₁ * P₂) V₂ = (0.998 atm * 2.1 L * 301 K) / (309 K * 0.900 atm) V₂ = 2.268 L

Therefore, the new volume of the balloon when it reaches the elevation with a pressure of 0.900 atm and a temperature of 28°C will be 2.268 L. This demonstrates how changes in temperature and pressure affect the volume of a gas according to the Ideal Gas Law.