20 grams of ethane (c2h6) in a 15 l container has a pressure of 1.3 bar. what is the temperature, in oc

The temperature is 42.5 °C

Explanation:

We apply the Law of Ideal Gases to solve this:

P  .  V = n .  R . T

First, we convert the bar into atm, so we make a rule of three.

1.013 bar is 1 atm

1.3 bar is (1.3 . 1) /1.013 = 1.28 atm

1.28atm . 15L = n . 0.082 . T

We must convert the mass to moles ( mass / molar mass)

20 g / 30 g / mol = 0.666 moles

1.28atm . 15L = 0.666 mol . 0.082 . T

(1.28 atm . 15L) / (0.666 mol . 0.082) = T

315.5 K = T

As this is absolute temperature we must convert to °C

315.5 K - 273= 42.5 °C

The temperature is 79.52 °C

Explanation:

Step 1: Data given

Mass of ethane = 20.0  grams

Volume of the gas = 15.0 L

Pressure in the container = 1.3 bar = 1.283 atm

Molar mass of ethane = 30.07 g/mol

Step 2: Calculate moles of ethane

moles ethane = mass ethane / molar mass ethane

Moles ethane = 20.0 grams / 30.07 g/mol

Moles ethane = 0.665 moles

Step 3: Calculate temperature

p*V = n*R*T

⇒ p = the pressure in the container = 1.283 atm

⇒ V = the volume = 15.0 L

⇒ n = the moles of ethane = 0.665 moles

⇒ R = the gas constant = 0.08206 L*atm/mol*K

⇒ T = the temperature = TO BE DETERMINED

T = (p*V)/(n*R)

T = (1.283*15)/(0.665*0.08206)

T = 352.67 K

T = 79.52 °C

The temperature is 79.52 °C

what are the two models

explanation: