The air pressure in a tire is 1.20 atm at 77.0 C. Find the pressure of the same amount of gas in the same volume at -12.0 C
(P1/T1) = (P2/T2)
To find the pressure of the same amount of gas in the same volume at a different temperature, we can use the combined gas law equation, which combines Boyle's Law, Charles's Law, and Gay-Lussac's Law. The equation is as follows:
(P₁ × V₁) / (T₁) = (P₂ × V₂) / (T₂)
Where:
P₁ and P₂ are the initial and final pressures, respectively.
V₁ and V₂ are the initial and final volumes, respectively.
T₁ and T₂ are the initial and final temperatures, respectively.
Let's use this equation to solve the problem:
Given:
P₁ = 1.20 atm
T₁ = 77.0 °C = 77.0 + 273.15 K (convert to Kelvin)
To convert Celsius to Kelvin, we add 273.15 to the given temperature. So, T₁ = 350.15 K.
We need to find:
P₂ (pressure at -12.0 °C)
Now we can set up the equation using the given values:
(1.20 atm × V₁) / (350.15 K) = (P₂ × V₂) / (261.15 K)
Since the volume (V₁ = V₂) is constant, we can cancel it from both sides of the equation:
1.20 atm / 350.15 K = P₂ / 261.15 K
Now solve for P₂:
P₂ = (1.20 atm / 350.15 K) × 261.15 K
P₂ ≈ 0.895 atm
Therefore, the pressure of the same amount of gas in the same volume at -12.0 °C is approximately 0.895 atm.