High School

A gas occupies 35.8 L at 1.13 atm and 42.3 °C. What would be the temperature of this gas, in Celsius, if the volume were decreased to 25.0 L and the pressure was left unchanged?

Answer :

Final answer:

Using Charles's Law, which establishes a relationship between the volume and absolute temperature of a gas under constant pressure, the final temperature of the gas when the volume is reduced to 25.0L is approximately -52.65°C.

Explanation:

This question is about the principles of gas laws, specifically Charles's Law which states: the volume of a gas is directly proportional to its temperature, provided the pressure and the amount of gas remain constant. In the current case, pressure is constant at 1.13 atm. The initial temperature is 42.3°C, but since we work with the Kelvin scale in gas law equations, we convert it to about 315.45K (by adding 273.15).

Now, using Charles's Law, V1/T1 = V2/T2 , where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature. From the problem, V1 is 35.8L, T1 is 315.45K, and V2 is 25.0L. We can solve this equation for T2 to find:

T2 = V2 * (T1/V1) = 25.0L * (315.45K / 35.8L)

This calculation results in T2 = 220.5K. Then, convert this value back to Celsius, subtracting 273.15, giving a final temperature of -52.65°C.

Learn more about Charles's Law here:

https://brainly.com/question/16927784

#SPJ11