Question

At what temperature would 0.500 moles of gas particles stored in a 100.0 mL container reach a pressure of 15.0 atm?

Answer 1

Hey there!

In this case, the equation of Clapeyron is used :

R = 0.082 

Volume in liters :

100.0 mL / 1000 => 0.1 L 

P * V = n * R * T

15.0 * 0.1 = 0.500 * 0.082 * T

1.5 = 0.041 * T

T = 1.5 / 0.041

T = 36.5 K

Answer C

Answer 2

Answer: Option (C) is the correct answer.

Explanation:

According to ideal gas law, product of pressure and volume equals n times R times T.

Mathematically,       PV = nRT

where         P = pressure

                  V = volume

                  n = number of moles

                  R = gas constant

                  T = temperature

Since it is known that value R = 0.082 L $atm mol^{-1} K^{-1}$ and the other values are given as P = 15.0 atm, V = 100 mL = 0.1 L, and n = 0.5 moles.

Therefore, calculate value of temperature as follows.

                           PV = nRT

                $15 atm \times 0.1 L$ = $0.5 moles \times 0.082 L atm mol^{-1} K^{-1} \times T$

               T = 36.58 K

Thus, we can conclude that temperature is 36.5 K.