A balloon under constant pressure goes from a very cold freezer at 150 K to a hot car at 300 K.
A. Volume is cut in half
B. Volume doubles
C. Volume stays the same
D. Volume increases from 150 k
C. Volume stays the same
To determine what happens to the volume of the balloon as it goes from a very cold freezer at 150 K to a hot car at 300 K, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
Since the balloon is under constant pressure, we can rearrange the ideal gas law equation to solve for the volume using V = nRT/P.
Considering that the pressure is constant, and assuming that the number of moles of gas and the ideal gas constant remain the same, we can focus on the relationship between the temperature and the volume.
Using the equation V = nRT/P, we can see that when the temperature is doubled, the volume will also double if everything else remains constant. Conversely, if the temperature is halved, the volume will also halve.
In this case, the temperature increases from 150 K to 300 K, so the volume of the balloon will increase accordingly. Therefore, the correct answer is:
D. Volume increases from 150 K.
To determine how the volume of a balloon under constant pressure changes when it goes from a freezer at 150 K to a hot car at 300 K, we can use the ideal gas law.
The ideal gas law is given by the equation PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
When the pressure is constant, the equation simplifies to V/T = constant.
In this scenario, we can assume that the number of moles of gas remains constant, as there is no mention of adding or removing gas from the balloon.
Given that the initial temperature in the freezer is 150 K and the final temperature in the hot car is 300 K, we can compare the ratio of the initial and final temperatures to determine how the volume changes.
V1/T1 = V2/T2
V1/150 = V2/300
Now let's consider the options:
A. Volume is cut in half: This option implies that V2 = V1/2. However, based on the equation above, this would only be true if the temperature ratio was 2:1, which is not the case here. So, option A is not correct.
B. Volume doubles: This option implies that V2 = 2V1. Based on the equation above, this would only be true if the temperature ratio was 1:2, which is the case here. So, option B is correct.
C. Volume stays the same: This option implies that V1 = V2. Based on the equation above, this would only be true if the temperature ratio was 1:1, which is not the case here. So, option C is not correct.
D. Volume increases from 150 K: This option doesn't provide enough information to determine the relationship between V1 and V2, as it only states the initial temperature. So, option D is not correct.
Therefore, the correct answer is B. The volume of the balloon doubles when it goes from a very cold freezer at 150 K to a hot car at 300 K under constant pressure.