Answer:
0.89L ( 2 decimal place)
Explanation:
Using Combined Gas law.
( P₁V₁)/T₁ = (P₂V₂) / T₂
P₁ = 1.00 atm
V₁ = 1.55L
T₁ = 27.0 °C = converting to kelvin ( 27 + 273k) = 300k
P₂ = same as P₁ = 1.00 atm
V₂ = ?
T₂ = -100°C = converting to kelvin ( -100 + 273k) = 173k
(1 x 1.55)/300 = (1 x V)/173
1.55/300 = V/173
V / 173 = 0.00516666666
V = 173 x 0.00516666666 = 0.89383333333 ≈ 0.89L ( 2 decimal place)
Answer:
The volume at -100 °C is 0.894 L
Explanation:
Step 1: Data given
The initial volume = 1.55L
The initial temperature = 27.0 °C = 300 K
The pressure is 1.00 atm and stays constant
The temperature lowers to -100 °C = 173 K
Step 2: Calculate the volume
V1 / T1 = V2 / T2
⇒ with V1 = the initial volume = 1.55L
⇒ with T1 = the initial temperature = 300 K
⇒ with V2 = The new volume = TO BE DETERMINED
⇒ with T2 = the final temperature = 173 K
1.55 / 300 = V2 /173
V2 = 0.894 L
The volume at -100 °C is 0.894 L
