Question

The gas left in a used aerosol can is at a pressure of 103 kPa at 25.0 °C. If the can heats up to 50.0 °C, what is the pressure of the gas inside the can, assuming the volume of the can is constant?

Answer 1

The question mentions a change in temperature from 25 to 50 °C. With that, the aim of the question is to determine the change in volume based on that change in temperature. Therefore this question is based on Gay- Lussac's Gas Law which notes that an increase in temperature, causes an increase in pressure since the two are directly proportional (once volume remains constant). Thus Gay-Lussac's Equation can be used to solve for the answer.
Boyle's Equation:
`(P_(1) )/(T_(1) )` =
`(P_(2) )/(T_(2) )`
Since the initial temperature (T₁) is 25 C, the final temperature is 50 C (T₂) and the initial pressure (P₁) is 103 kPa, then we can substitute these into the equation to find the final pressure (P₂).

`(P_(1) )/(T_(1) )` =
`(P_(2) )/(T_(2) )`∴ by substituting the known values, ⇒ (103 kPa) ÷ (25 °C) = (P₂) ÷ (50 °C)
⇒ P₂ = (4.12 kPa · °C) (50 °C)
= 206 kPa
Thus the pressure of the gas since the temperature was raised from 25 °C to 50 °C is 206 kPa