Aerosol cans carry clear warnings against incineration because of the high pressures that can develop upon heating. Suppose a can contains a residual amount of gas at a pressure of 755 mm Hg and a temperature of 25 °C. What would the pressure be if the can were heated to 1155 °C?
Well, if you heat an aerosol can to 1155 °C, you might not just have to worry about the pressure, but also about melting your entire kitchen! But let's focus on the pressure for now.
To solve this problem, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
First, let's convert the initial temperature of 25 °C to Kelvin by adding 273.15:
25 °C + 273.15 = 298.15 K
We also need to convert the final temperature of 1155 °C to Kelvin:
1155 °C + 273.15 = 1428.15 K
Now, let's plug the values into the ideal gas law equation:
P1V1/T1 = P2V2/T2
Since the volume and number of moles are constant, we can simplify the equation to:
P1/T1 = P2/T2
Now, substitute the values:
(755 mm Hg)/(298.15 K) = P2/(1428.15 K)
Now, cross multiply and solve for P2:
P2 = (755 mm Hg) * (1428.15 K) / (298.15 K)
Calculating this, we find:
P2 ≈ 3613 mm Hg
So, if you heat the can to a scorching 1155 °C, the pressure inside would be approximately 3613 mm Hg. But, as a friendly reminder, don't actually try this at home! Keep those aerosol cans away from excessive heat to avoid any unexpected pyrotechnics!
To determine the pressure if the can were heated to 1155 °C, we can use the ideal gas law:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
Since we're only interested in the change in pressure, we can set up the following equation:
P1/T1 = P2/T2
Where:
P1 = initial pressure (755 mm Hg)
T1 = initial temperature (25 °C + 273.15 K)
P2 = final pressure (unknown)
T2 = final temperature (1155 °C + 273.15 K)
Now let's substitute the values into the equation:
(755 mm Hg)/(25 °C + 273.15 K) = P2/(1155 °C + 273.15 K)
Calculating the values:
Temperature in Kelvin:
T1 = 25 °C + 273.15 K = 298.15 K
T2 = 1155 °C + 273.15 K = 1428.15 K
P2 = (755 mm Hg) * (1428.15 K) / (298.15 K)
Simplifying the equation:
P2 ≈ 3612.25 mm Hg
Therefore, the pressure in the can would be approximately 3612.25 mm Hg if it were heated to 1155 °C.
To solve this problem, you can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
To find the pressure at 1155 °C, you need to convert the temperature to Kelvin. Kelvin is the absolute temperature scale and is obtained by adding 273.15 to the temperature in Celsius. So, 25 °C + 273.15 = 298.15 K.
Now, let's calculate the pressure at 298.15 K using the ideal gas law:
P1V1 = nRT1
Since the can is sealed, the number of moles and the volume remain constant. Therefore, we can write:
P1/T1 = P2/T2
Where P1 is the initial pressure (755 mm Hg), T1 is the initial temperature (298.15 K), P2 is the final pressure (what we want to find), and T2 is the final temperature (1155 °C + 273.15 = 1428.15 K).
Now, we can rearrange the equation to solve for P2:
P2 = (P1 x T2) / T1
Substituting the values we know:
P2 = (755 mm Hg x 1428.15 K) / 298.15 K
Calculating this gives us:
P2 ≈ 3611.92 mm Hg
Therefore, if the can were heated to 1155 °C, the pressure would be approximately 3611.92 mm Hg. It's important to note that the pressure value obtained here is significantly higher than typical atmospheric pressure, which highlights the dangers of incinerating aerosol cans.