Respuesta :
Answer:
The final volume of the sample of gas [tex]V_{2}[/tex] = 0.000151 [tex]m^{3}[/tex]
Explanation:
Initial volume [tex]V_{1}[/tex] = 200 ml = 0.0002 [tex]m^{3}[/tex]
Initial temperature [tex]T_{1}[/tex] = 296 K
Initial pressure [tex]P_{1}[/tex] = 101.3 K pa
Final temperature [tex]T_{2}[/tex] = 336 K
Final pressure [tex]P_{2}[/tex] = K pa
Relation between P , V & T is given by
[tex]P_{1} \frac{V_{1} }{T_{1} } = P_{2} \frac{V_{2} }{T_{2} }[/tex]
Put all the values in the above equation we get
[tex]101.3 (\frac{0.0002}{296} )= 152 (\frac{V_{2} }{336} )[/tex]
[tex]V_{2}[/tex] = 0.000151 [tex]m^{3}[/tex]
This is the final volume of the sample of gas.
Taking into account the combined law equation, the volume of the sample is 151.3 mL.
- Charles's Law
First of all, you have to know that Charles's Law consists of the relationship that exists between the volume and the temperature of a certain quantity of ideal gas.
This law says that when the amount of gas and pressure are kept constant, the quotient between the volume and the temperature will always have the same value:
[tex]\frac{V}{T}=k[/tex]
- Boyle's Law
Boyle's law is one of the gas laws that relates the volume and pressure of a certain quantity of gas kept at constant temperature. Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container.
Boyle's law is expressed mathematically as:
P×V= k
- Gay-Lussac's Law
Gay-Lussac's law is one of the gas laws that establishes the relationship between the temperature and the pressure of a gas when the volume is constant.
Gay-Lussac's law is expressed mathematically as:
[tex]\frac{P}{T}=k[/tex]
- Combined law equation
Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:
[tex]\frac{PxV}{T}=k[/tex]
Studying two different states, an initial state 1 and a final state 2, it is satisfied:
[tex]\frac{P1xV1}{T1}=\frac{P2xV2}{T2}[/tex]
- This case
In this case you know:
- P1= 101.3 kPa
- V1= 200 mL
- T1= 296 K
- P2= 152 kPa
- V2= ?
- T2= 336 K
Replacing in combined law equation:
[tex]\frac{101.3 kPax200 mL}{296 K}=\frac{152 kPaxV2}{336 K}[/tex]
Solving:
[tex]\frac{101.3 kPax200 mL}{296 K}x\frac{336 K}{152 kPa} =V2[/tex]
V2= 151.3 mL
Finally, the volume of the sample is 151.3 mL.
Learn more:
- https://brainly.com/question/4147359?referrer=searchResults
