A volume of 10.0L of gas at a temperature of 5c is cooled to a temperature of 85C at constant pressure what is the new volume of
gas
2 answers:
Answer:
At constant pressure the new volume of gas is 12.878 L.
Explanation:
Charles's Law consists of the relationship between the volume and temperature of a certain amount of ideal gas, which is maintained at a constant pressure, by means of a proportionality constant that is applied directly. So the ratio between volume and temperature will always have the same value:
where the temperature is expressed in degrees kelvin (° K)
Then, when considering the two situations 1 and 2, keeping the amount of gas and the temperature constant, the relationship must be met:
In this case, you know:
- V1= 10 L
- T1= 5 +273 = 278°K (where 0°C=273°K)
- V2= ?
- T2= 85 + 273 = 358°K
Replacing:
Solving:
V2=12.878 L
<u><em>At constant pressure the new volume of gas is 12.878 L.</em></u>
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The question is incomplete, the complete question is attached below.
Answer : The volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.
Explanation : Given,
Mass of NaOH = 60 g
Volume of stock solution = 300 mL
Molar mass of NaOH = 40 g/mol
First we have to calculate the molarity of stock solution.
Now put all the given values in this formula, we get:
Now we have to determine the volume of stock solution and distilled water mixed.
Formula used :
where,
are the molarity and volume of stock solution.
are the molarity and volume of diluted solution.
From data (A) :
Putting values in above equation, we get:
Volume of stock solution = 20 mL
Volume of distilled water = 100 mL - 20 mL = 80 mL
From data (B) :
Putting values in above equation, we get:
Volume of stock solution = 20 mL
Volume of distilled water = 100 mL - 20 mL = 80 mL
From data (C) :
Putting values in above equation, we get:
Volume of stock solution = 60 mL
Volume of distilled water = 300 mL - 60 mL = 240 mL
From data (D) :
Putting values in above equation, we get:
Volume of stock solution = 60 mL
Volume of distilled water = 300 mL - 60 mL = 240 mL
From this we conclude that, when 20 mL stock solution and 80 mL distilled water mixed then it will result in a solution that is approximately 1 M NaOH.
Hence, the volumes of stock solution and distilled water will be, 20 mL and 80 mL respectively.
