A basketball with a volume of 3.50 Liters is left in a car when the temperature is 298 K. What will the volume of the ball be as the car heats up to 305 K during the day? (2.0 p
(V1/T1) = (V2/T2)
To solve this problem, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature, assuming constant pressure.
The equation for Charles's Law is:
V1/T1 = V2/T2
where V1 and V2 are the initial and final volumes, and T1 and T2 are the initial and final temperatures.
Given:
V1 = 3.50 L
T1 = 298 K
T2 = 305 K
Let's plug in the values and solve for V2:
V1/T1 = V2/T2
3.50 L / 298 K = V2 / 305 K
Now we'll cross-multiply and solve for V2:
V2 = (3.50 L / 298 K) * 305 K
V2 = 3.61 L
So, the volume of the ball when the car heats up to 305 K will be approximately 3.61 L.
To solve this problem, we can use Charles' Law, which states that the volume of a gas is directly proportional to its absolute temperature, assuming constant pressure.
The formula for Charles' Law is:
V1/T1 = V2/T2
Where:
V1 = initial volume (3.50 Liters)
T1 = initial temperature (298 K)
V2 = final volume (unknown)
T2 = final temperature (305 K)
We need to rearrange the equation to solve for V2:
V2 = (V1 x T2) / T1
Now we can substitute the given values into the formula:
V2 = (3.50 Liters x 305 K) / 298 K
V2 = (1067.50 Liters * K) / K
V2 ≈ 1067.50 Liters
Therefore, the volume of the basketball as the car heats up to 305 K will be approximately 1067.50 Liters.