A 10,0-L cylinder of gas is stored at room temperature (20.0°C) and a pressure of 1800 psi. If the gas is
1 answer:
Considering the Charles' law, the gas would have a temperature of -109.2 C.
<h3>Charles' law</h3>Finally, Charles' law establishes the relationship between the volume and temperature of a gas sample at constant pressure. This law says that the volume is directly proportional to the temperature of the gas. That is, if the temperature increases, the volume of the gas increases, while if the temperature of the gas decreases, the volume decreases.
Charles' law is expressed mathematically as:
If you want to study two different states, an initial state 1 and a final state 2, the following is true:
In this case, you know:
- P1= 1800 psi
- V1= 10 L
- T1= 20 C= 293 K (being 0 C= 273 K)
- P2= 1800 psi
- V2= 6 L
- T2= ?
You can see that the pressure remains constant, so you can apply Charles's law.
Replacing in the Charles's law:
Solving:
<u><em>T2=163.8 K= -109.2 C</em></u>
The gas would have a temperature of -109.2 C.
Learn more about Charles's law:
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Answer:but-1-ene
Explanation:This is an E2 elimination reaction .
Kindly refer the attachment for complete reaction and products.
Sodium tert-butoxide is a bulky base and hence cannot approach the substrate 2-chlorobutane from the more substituted end and hence major product formed here would not be following zaitsev rule of elimination reaction.
Sodium tert-butoxide would approach from the less hindered side that is through the primary centre and hence would lead to the formation of 1-butene .The major product formed in this reaction would be 1-butene .
As the mechanism of the reaction is E-2 so it will be a concerted mechanism and as sodium tert-butoxide will start abstracting the primary hydrogen through the less hindered side simultaneously chlorine will start leaving. As the steric repulsion in this case is less hence the transition state is relatively stabilised and leads to the formation of a kinetic product 1-butene.
Kinetic product are formed when reactions are dependent upon rate and not on thermodynamical stability.
2-butene is more thermodynamically6 stable as compared to 1-butene
The major product formed does not follow the zaitsev rule of forming a more substituted alkene as sodium tert-butoxide cannot approach to abstract the secondary proton due to steric hindrance.
