The new pressure is 2 Kpa
<em><u>Solution:</u></em>
At constant temperature, the pressure of a sample of gas is inversely proportional to its volume
Therefore,
![\text{ pressure of a sample of gas} \propto \frac{1}{\text{volume}}](https://tex.z-dn.net/?f=%5Ctext%7B%20pressure%20of%20a%20sample%20of%20gas%7D%20%5Cpropto%20%5Cfrac%7B1%7D%7B%5Ctext%7Bvolume%7D%7D)
Let "p" be the pressure of sample gas
Let "v" be the volume
Then we get,
![p \propto \frac{1}{v}\\\\p = k \times \frac{1}{v}\\\\p = \frac{k}{v} ---------- eqn 1](https://tex.z-dn.net/?f=p%20%5Cpropto%20%5Cfrac%7B1%7D%7Bv%7D%5C%5C%5C%5Cp%20%3D%20k%20%5Ctimes%20%5Cfrac%7B1%7D%7Bv%7D%5C%5C%5C%5Cp%20%3D%20%5Cfrac%7Bk%7D%7Bv%7D%20----------%20eqn%201)
Where, "k" is the constant of proportionality
<em><u>I have some oxygen in a 2.28 liter container with a pressure of 5 kPa</u></em>
Substitute v = 2.28 and p = 5 in eqn 1
![5 = \frac{k}{2.28}\\\\k = 5 \times 2.28\\\\k = 11.4](https://tex.z-dn.net/?f=5%20%3D%20%5Cfrac%7Bk%7D%7B2.28%7D%5C%5C%5C%5Ck%20%3D%205%20%5Ctimes%202.28%5C%5C%5C%5Ck%20%3D%2011.4)
<em><u>I move all of it to a 5.7 liter container at the same temperature</u></em>
Substitute k = 11.4 and v = 5.7 in eqn 1
![p = \frac{11.4}{5.7}\\\\p = 2](https://tex.z-dn.net/?f=p%20%3D%20%5Cfrac%7B11.4%7D%7B5.7%7D%5C%5C%5C%5Cp%20%3D%202)
Thus new pressure is 2 kPa
Answer:
hi! your answer should be D
Step-by-step explanation:
I plugged the equation into a calculator and got D.
Answer:
sddssddsdssdsd
Step-by-step explanation:
sddsdsdsdsds
Answer:
aₙ = 31n - 5
Step-by-step explanation:
The terms of your sequence are.
26, 57, 88, …
This is an arithmetic sequence, because there is a constant difference of 31 between consecutive terms.
The explicit formula for the nth term of an arithmetic sequence is
aₙ = a₁ + d(n - 1 )
a₁ is the first term, and d is the difference in value between consecutive terms. Thus,
aₙ = 26 + 31(n - 1) = 26 + 31n - 31
aₙ = 31n - 5
The explicit rule for the arithmetic sequence is aₙ = 31n - 5.