An air bubble has a volume of 0.500 L at 18 C. If the pressure does not change, what is the volume, in liters, at each of the following temperatures?
A. 0 C
B. 425 K
C. -12 C
D. 575 K
what is the answer?!
To answer this question, we need to use the ideal gas law, which states that the pressure (P) times the volume (V) is equal to the number of moles (n) times the gas constant (R) times the temperature (T). The equation is written as:
PV = nRT
Given that the pressure does not change, we can assume that the number of moles and gas constant also remain constant. Therefore, we can use the equation to find the volume at different temperatures.
We can rewrite the equation as:
V₁/T₁ = V₂/T₂
Where V₁ is the initial volume (0.500 L) and T₁ is the initial temperature (18 °C).
Let's calculate the volume at each of the given temperatures:
A. 0 °C:
Converting 0 °C to Kelvin: T₁ = 0 + 273 = 273 K
Using the equation, we can find V₂:
V₂ = (V₁/T₁) * T₂
V₂ = (0.500 L / 291 K) * 273 K
V₂ = 0.464 L
Therefore, the volume of the air bubble at 0 °C would be approximately 0.464 L.
B. 425 K:
Using the equation, we can find V₂:
V₂ = (0.500 L / 291 K) * 425 K
V₂ = 0.732 L
Therefore, the volume of the air bubble at 425 K would be approximately 0.732 L.
C. -12 °C:
Converting -12 °C to Kelvin: T₁ = -12 + 273 = 261 K
Using the equation, we can find V₂:
V₂ = (0.500 L / 291 K) * 261 K
V₂ = 0.450 L
Therefore, the volume of the air bubble at -12 °C would be approximately 0.450 L.
D. 575 K:
Using the equation, we can find V₂:
V₂ = (0.500 L / 291 K) * 575 K
V₂ = 0.986 L
Therefore, the volume of the air bubble at 575 K would be approximately 0.986 L.
To solve this problem, we can use Charles's law, which states that the volume of a gas is directly proportional to its temperature (in Kelvin) at constant pressure. The equation for Charles's law is:
V1/T1 = V2/T2
Where:
V1 = initial volume
T1 = initial temperature
V2 = final volume (unknown)
T2 = final temperature (unknown)
A. To find the volume at 0 °C, we need to convert the temperature to Kelvin.
0 °C = 273.15 K
We'll plug the values into the equation:
0.500 L / (18 °C + 273.15 K) = V2 / (0 °C + 273.15 K)
Simplifying the equation:
0.500 L / 291.15 K = V2 / 273.15 K
Cross multiply and solve for V2:
V2 = (0.500 L * 273.15 K) / 291.15 K = 0.468 L
Therefore, the volume at 0 °C is 0.468 L.
B. To find the volume at 425 K, we'll plug the values into the equation:
0.500 L / (18 °C + 273.15 K) = V2 / 425 K
Cross multiply and solve for V2:
V2 = (0.500 L * 425 K) / (18 °C + 273.15 K) = 0.512 L
Therefore, the volume at 425 K is 0.512 L.
C. To find the volume at -12 °C, we'll convert the temperature to Kelvin:
-12 °C = 261.15 K
We'll plug the values into the equation:
0.500 L / (18 °C + 273.15 K) = V2 / ( -12 °C + 273.15 K)
Cross multiply and solve for V2:
V2 = (0.500 L * (-12 °C + 273.15 K)) / (18 °C + 273.15 K) = 0.469 L
Therefore, the volume at -12 °C is 0.469 L.
D. To find the volume at 575 K, we'll plug the values into the equation:
0.500 L / (18 °C + 273.15 K) = V2 / 575 K
Cross multiply and solve for V2:
V2 = (0.500 L * 575 K) / (18 °C + 273.15 K) = 0.787 L
Therefore, the volume at 575 K is 0.787 L.