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?
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To solve this problem, we can use the ideal gas law, which relates pressure (P), volume (V), and temperature (T) of a gas:
PV = nRT
where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of gas
R is the ideal gas constant
T is the temperature of the gas in Kelvin
First, we need to convert the given temperatures to Kelvin. The Kelvin temperature scale is used in gas calculations as it does not have negative values. To convert Celsius to Kelvin, we use the equation:
T(K) = T(°C) + 273.15
Given:
Initial temperature (T1) = 25 °C = 25 + 273.15 K = 298.15 K
Final temperature (T2) = 1155 °C = 1155 + 273.15 K = 1428.15 K
Now, we need to find the ratio of the initial pressure to volume (P1/V1). Since the amount of gas (moles) and the volume are constant, this ratio remains the same:
P1/V1 = P2/V2
Substituting the given values:
P1 = 755 mm Hg
V1 = Unknown (we do not need the volume in this problem)
We are looking for the final pressure (P2) when the temperature is increased to T2 = 1428.15 K.
Now we need to rearrange the equation to solve for P2:
P2 = P1 * (V2 / V1)
Since we do not need the volume in this problem, we can eliminate it by taking the ratio of temperatures:
P2 = P1 * (T2 / T1)
Substituting the given values:
P1 = 755 mm Hg
T1 = 298.15 K
T2 = 1428.15 K
P2 = 755 mm Hg * (1428.15 K / 298.15 K)
Now we can calculate P2:
P2 = 755 * (1428.15 / 298.15)
P2 ≈ 3619.95 mm Hg
Therefore, if the can were heated to 1155 °C, the pressure inside the can would be approximately 3619.95 mm Hg.