To solve this problem, we can use the combined gas law, which states:
(P₁ × V₁) / (T₁ × n₁) = (P₂ × V₂) / (T₂ × n₂)
where: P₁ and P₂ are the initial and final pressures respectively, V₁ and V₂ are the initial and final volumes respectively (assumed constant), T₁ and T₂ are the initial and final temperatures respectively, n₁ and n₂ are the initial and final moles of gas respectively.
Since the problem states that the can is filled with an ideal gas, we can assume that the number of moles of gas remains constant (n₁ = n₂). We can also assume that the volume of the can remains constant, so (V₁ = V₂). Therefore, the equation simplifies to:
(P₁ / T₁) = (P₂ / T₂)
Let's convert the temperatures to Kelvin:
T₁ = 20.0 °C + 273.15 = 293.15 K T₂ = 50.0 °C + 273.15 = 323.15 K
Now, we can rearrange the equation to solve for P₂:
P₂ = (P₁ × T₂) / T₁
Substituting the given values: P₁ = 573 kPa T₁ = 293.15 K T₂ = 323.15 K
P₂ = (573 kPa × 323.15 K) / 293.15 K
P₂ ≈ 631.22 kPa
To convert the pressure to atmospheres (ATM), we can use the conversion factor:
1 kPa ≈ 0.00987 ATM
P₂ ≈ 631.22 kPa × 0.00987 ATM/kPa
P₂ ≈ 6.23 ATM
Therefore, the pressure in the aluminum aerosol can at 50.0 °C would be approximately 6.23 ATM.