What happens to the average kinetic energy of the molecules of a warm substance when the temperature of the substance is slightl
2 answers:
Answer: Option (b) is the correct answer.
Explanation:
Average kinetic energy is defined as half of mass of gas times the square of root mean square velocity.
Mathematically, Average K.E =
Also, Kinetic energy =
So, when there is increase in temperature then there will be increase in movement of the molecules. As a result, kinetic energy will also increase. Since, kinetic energy is proportional to temperature.
Therefore, with decrease in temperature there will be decrease in average kinetic energy of the molecules.
Thus, we can conclude that the average kinetic energy of the molecules of a warm substance also decreases when the temperature of the substance is slightly decreased.
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Answer:
Explanation:
Hello there!
In this case, according to the given information, it turns out possible for us to calculate the required new volume by using the Charles' law as a directly proportional relationship between temperature and volume:
In such a way, we solve for V2 and plug in V1, T1 and T2 to obtain:
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Answer:
B.3/5p
Explanation:
For this question, we have to remember <u>"Dalton's Law of Partial Pressures"</u>. This law says that the pressure of the mixture would be equal to the sum of the partial pressure of each gas.
Additionally, we have a <em>proportional relationship between moles and pressure</em>. In other words, more moles indicate more pressure and vice-versa.
Where:
=Partial pressure
=Total pressure
=mole fraction
With this in mind, we can work with the moles of each compound if we want to analyze the pressure. With the molar mass of each compound we can calculate the moles:
<u>moles of hydrogen gas</u>
The molar mass of hydrogen gas () is 2 g/mol, so:
<u>moles of oxygen gas</u>
The molar mass of oxygen gas () is 32 g/mol, so:
Now, total moles are:
Total moles = 2 + 3 = 5
With this value, we can write the partial pressure expression for each gas:
So, the answer would be <u>3/5P</u>.
I hope it helps!