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?

A) 36500 K
B) 7.96 K
C) 36.5 K
D) 3.60 K

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.

Step-by-step 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 * 0.1 L` =
`0.5 moles * 0.082 L atm mol^(-1) K^(-1) * T`

T = 36.58 K

Thus, we can conclude that temperature is 36.5 K.