Respuesta :
Use Charles’ Law: V1/T1 = V2/T2. Given a mass of gas held at a constant pressure, changes in the gas’s temperature and volume are directly proportional. Since the volume of the gas in the balloon decreased, we should expect that the temperature had also decreased.
Here, V1 = 12.0 L, T1 = 357.15 K, and V2 = 5.0 L. To solve for T2, we rearrange the equation and compute:
T2 = V2T1/V1 = (5.0 L)(357.15 K)/(12.0 L)
T2 = 148.8125 K - 273.15 = -124.34 °C
To two significant figures, the answer would be -120 °C.
Here, V1 = 12.0 L, T1 = 357.15 K, and V2 = 5.0 L. To solve for T2, we rearrange the equation and compute:
T2 = V2T1/V1 = (5.0 L)(357.15 K)/(12.0 L)
T2 = 148.8125 K - 273.15 = -124.34 °C
To two significant figures, the answer would be -120 °C.
The ideal gas relates the volume, pressure, and temperature to each other. The balloon was cooled to -120 degrees celsius.
What is Charle's law?
Charle's law states the direct relationship between the volume and the temperature of the hypothetical gas. The law is given as,
[tex]\rm \dfrac{V_{1}}{T_{1}} = \rm \dfrac{V_{2}}{T_{2}}[/tex]
As it can be seen from the question the volume of the gas in the balloon is decreasing so the temperature will also decrease.
Given,
Initial volume = 12.0 L
Final volume = 5.0 L
Initial temperature = 357.15 K
The final temperature is calculated as:
[tex]\begin{aligned} \rm T_{2} &= \rm \dfrac{V_{2}T_{1}}{V_{1}}\\\\&= \dfrac{5\times 357.15}{12}\\\\&= 148.81\;\rm K\end{aligned}[/tex]
Therefore, -120 degrees celsius is the final temperature of the balloon.
Learn more about Charle's law here:
https://brainly.com/question/16404056
#SPJ2
