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8. What is the pressure in atm inside a 3.00 L ball that contains 1.23 moles of air at 32 degrees C?

Respuesta :

lmaisunny

Answer:

P = 10.3 atm

Explanation:

n = 1.23

V = 3.00 L

T = 32C = 305.15K

P = ?

Follow this equation below

PV = nRT

P x 3.00 = 1.23 x 0.08206 x 305.15K

P x 3.00 = 30.8

Dividing both sides by 3.00

P = 10.3 atm

Hope this helps!!

The pressure inside a 3.00 L ball that contains 1.23 moles of air at 32 degrees C is 10.25 atm.

How do we calculate the pressure?

Pressure of gas will be calculated by using the ideal gas equation as:

PV = nRT, where

  • P = pressure = ?
  • V = volume = 3L
  • n = moles = 1.23 mol
  • R = universal gas constant = 0.082 L.atm/K.mol
  • T = temperature = 32 degrees C = 305 K

On putting all values, we get

P = (1.23)(0.082)(305) / (3) = 10.25 atm

Hence required pressure is 10.25 atm.

To know more about ideal gas equation, visit the below link:

https://brainly.com/question/12873752

#SPJ2

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