Answer:
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
Answer:
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
