Question

The gas in an aerosol can is at a pressure of 3.10 atm at 25 degrees Celsius. Directions on the can warn the user not to keep the can in a place above 52 degrees Celsius. What would the gas pressure in the can be at 52 degrees Celsius

Answer 1

`\boxed {\boxed {\sf 6.4 \ atm}}`

Step-by-step explanation:

We are asked to find the pressure of a gas in a can given a change in temperature. We will use Gay-Lussac's Law, which states the pressure of a gas is directly proportional to the temperature. The formula for this law is:

`\frac {P_1}{T_1}= \frac {P_2}{T_2}`

Initially, the gas in the aerosol can has a pressure of 3.10 atmospheres at a temperature of 25 degrees Celsius.

`\frac { 3.10 \ atm}{25 \textdegree C}=(P_2)/(T_2)`

The temperature is increased to 52 degrees Celsius, but the pressure is unknown.

`\frac { 3.10 \ atm}{25 \textdegree C}=(P_2)/(52 \textdegree C)`

We are solving for the new pressure, so we must isolate the variable
`P_2`. It is being divided by 52 degrees Celsius. The inverse operation of division is multiplication, so we multiply both sides of the equation by 52 °C.

`52 \textdegree C *\frac { 3.10 \ atm}{25 \textdegree C}=(P_2)/(52 \textdegree C) * 52 \textdegree C`

`52 \textdegree C *\frac { 3.10 \ atm}{25 \textdegree C}=P_2`

The units of degrees Celsius cancel.

`52 *\frac { 3.10 \ atm}{25}=P_2`

`52 *0.124 \ atm = P_2`

`6.448 \ atm = P_2`

The original values of pressure and temperature have 2 and 3 significant figures. Our answer must be rounded to the least number of sig figs, which is 2. For the number we calculated, that is the tenths place. The 4 in the hundredth place tells us to leave the 4 in the tenths place.

`6.4 \ atm \approx P_2`

The gas pressure in the can at 52 degrees Celsius is approximately 6.4 atmospheres.