If 3.13 mol of an ideal gas has a pressure of 2.33 atm and a volume of 72.31 L, what is the temperature of the sample in degrees Celsius?

382.49 C degree Celsius

Explanation:

Hello,

This problem deals with understanding the ideal gas law which hopes to predict how ideal gases might behave in any given condition. I listed the formula below and we are basically just going to solve for temperature by rearranging the equation as seen on the picture (there's also other rearranged ones in case you need to solve for those).

Universal gas constant R has a value of 0.0821 L * atm/(mole * K) when working with these given units so it will be part of this equation. R value changes based on what units you have.

T = PV/nR

   = (2.33) (72.31) / (3.13)(0.0821)

   = 655.64 K

Question is asking temperature in celsius so we employ the formula attached below:

C = K - 273.15

   = 655.64-273.15

    = 382.49 degree Celsius

382.49 degree Celsius is the answer!

answer

i think the answer is a) as air pressure in an area increases, the density of the gas particles in that area decreases.

hope it