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:     $\frac{P_{1} }{T_{1} }$    =   $\frac{P_{2} }{T_{2} }$
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).
                      $\frac{P_{1} }{T_{1} }$    =   $\frac{P_{2} }{T_{2} }$∴ 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