A balloon under constant pressure goes from a very cold freezer 150 K to a hot car at 300 K does the volume increase by 150 P does the volume double does the volume cut in half or does the volume stay the same?
Based on Charles's Law, which states that at constant pressure, the volume of a gas is directly proportional to its temperature, we can determine how the volume of the balloon changes.
The initial temperature of the balloon is 150 K, and the final temperature is 300 K. Since the temperatures are in Kelvin units, which is an absolute temperature scale, the change in temperature can be calculated as:
Change in temperature = Final temperature - Initial temperature
Change in temperature = 300 K - 150 K
Change in temperature = 150 K
If the volume of the balloon is directly proportional to the temperature, then the change in volume will also be proportional to the change in temperature. Since the change in temperature is 150 K, we can infer that the volume will increase by 150 P (units unspecified).
According to Charles's Law, the volume of a gas is directly proportional to its temperature at constant pressure. Therefore, if the pressure remains constant, the volume of the balloon will increase as the temperature increases.
To determine the exact change in volume, we can use the formula:
(V1 / T1) = (V2 / T2)
where V1 is the initial volume, T1 is the initial temperature, V2 is the final volume, and T2 is the final temperature.
Given that the initial temperature (T1) is 150 K and the final temperature (T2) is 300 K, and assuming the pressure remains constant, we can rearrange the equation to find the final volume (V2):
(V2 / 150 K) = (V1 / 300 K)
Multiplying both sides by 150 K:
V2 = (V1 / 300 K) * 150 K
As a result, the final volume (V2) will be proportional to the initial volume (V1) but decreased by a factor of 2 due to the increase in temperature. Therefore, the volume will be halved.
To determine how the volume of a balloon changes when it goes from a very cold freezer (150 K) to a hot car (300 K) under constant pressure, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (which remains constant)
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature in Kelvin
In this case, the pressure is constant, so we can simplify the equation to:
V1 / T1 = V2 / T2
Where:
V1 = initial volume in the freezer
T1 = initial temperature in the freezer
V2 = final volume in the hot car
T2 = final temperature in the hot car
Given that the volume V1 in the freezer is known, and the temperatures T1 and T2 are known, we can rearrange the equation to solve for V2:
V2 = (V1 * T2) / T1
Let's plug in the values given:
V1 = initial volume (unknown, we'll use a general value)
T1 = 150 K
T2 = 300 K
Now, let's analyze the options provided:
1. Does the volume increase by 150%?
To find out, we can calculate V2 using V1 as 100 units (just as an example):
V2 = (100 * 300) / 150 = 200
Since V2 is not 150% more than V1 (100 units), this option is not correct.
2. Does the volume double?
Let's calculate V2 using V1 as 100 units:
V2 = (100 * 300) / 150 = 200
Since V2 is exactly double V1, this option is correct.
3. Does the volume cut in half?
Again, let's calculate V2 using V1 as 100 units:
V2 = (100 * 300) / 150 = 200
Since V2 is not half of V1, this option is also not correct.
4. Does the volume stay the same?
Let's calculate V2 using V1 as 100 units:
V2 = (100 * 300) / 150 = 200
Since the volume V2 is not equal to V1 (100 units), this option is not correct.
Therefore, the correct answer is: the volume doubles when the balloon goes from the freezer to the hot car under constant pressure.