a basketgball is inflated to a pressure of 1,50 atm in a 20.c garage. What is the pressure of the basketball outside whre the temperature is -5.00 C?
(P1/T1) = (P2/T2)
Remember T must be in kelvin.
To solve this question, we can use the combined gas law, which relates the initial and final conditions of a gas sample. The combined gas law formula is expressed as:
(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.
Given that we have the initial and final temperatures and the initial pressure of the basketball, we can plug those values into the formula to find the final pressure outside the garage.
Step 1: Convert temperature to Kelvin
We need to convert the temperatures from Celsius to Kelvin since the gas law requires temperature to be in Kelvin scale.
Initial temperature (T1) in Kelvin:
T1 = 20°C + 273.15 = 293.15 K
Final temperature (T2) in Kelvin:
T2 = -5.00°C + 273.15 = 268.15 K
Step 2: Solve for the final pressure (P2)
Using the combined gas law formula, we have:
(P1 * V1) / T1 = (P2 * V2) / T2
Given:
P1 = 1.50 atm (initial pressure)
V1 = unknown (initial volume)
T1 = 293.15 K (initial temperature)
V2 = V1 (since the volume remains constant)
T2 = 268.15 K (final temperature)
The equation now becomes:
(1.50 atm * V1) / 293.15 K = (P2 * V1) / 268.15 K
To solve for P2, we multiply both sides of the equation by (268.15 K):
1.50 atm * V1 * 268.15 K / 293.15 K = P2 * V1
Simplifying:
P2 = (1.50 atm * 268.15 K) / 293.15 K
P2 ≈ 1.37 atm (rounded to two decimal places)
Therefore, the pressure of the basketball outside, where the temperature is -5.00°C, is approximately 1.37 atm.