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

Answer:

$\boxed {\boxed {\sf 6.4 \ atm}}$

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.

$\frac { 3.10 \ atm}{25 \textdegree C}=\frac{P_2}{T_2}$

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

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

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.

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

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

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.