Answer:
The answer to your question is: V2 = 4.2 l
Explanation:
Data
T1 = 330°K
P1 = 150 kPa
V1 = 3.5 l
STP
V2 = ?
T2 = 273°K
P2 = 101.3 kPa
Formula
[tex]\frac{P1V1}{T1} = \frac{P2V2}{T2}[/tex]
[tex]V2 = \frac{P1V1T2}{T1P2}[/tex]
Substitution
[tex]V2 = \frac{(150)(3.5)(273)}{(330)(101.3)}[/tex]
V2 = 143325 / 33429
Result V2 = 4.2 l
The volume of the sample at STP is 4.29 L
From the question, we are to determine the volume
From the combined gas law equation,
[tex]\frac{P_{1}V_{1} }{T_{1}} = \frac{P_{2}V_{2} }{T_{2}}[/tex]
Where P₁ is the initial pressure
V₁ is the initial volume
T₁ is the initial temperature
P₂ is the final pressure
V₂ is the final volume
T₂ is the final temperature
From the given information
P₁ = 150 KPa = 150000 Pa = 1.48 atm
T₁ = 330 K
V₁ = 3.50 L
At STP
Pressure = 1 atm and
Temperature = 273 K
Then,
P₂ = 1 atm
T₂ = 273.15 K
Putting the parameters into the formula, we get
[tex]\frac{1.48 \times 3.50}{330}=\frac{1\times V_{2} }{273.15}[/tex]
[tex]V_{2} = \frac{1.48 \times 3.50 \times 273.15}{1\times 330}[/tex]
[tex]V_{2} = 4.29 \ L[/tex]
Hence, the volume of the sample at STP is 4.29 L
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