What is the question that needs an answer?
Explanation:
Let us assume that the value of
= ![2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s](https://tex.z-dn.net/?f=2.82%20%5Ctimes%2010%5E%7B5%7D%20%5Ctimes%20e%5E%28%7B%5Cfrac%7B-9835%7D%7BT%7D%7D%29%20m%5E%7B6%7D%2Fmol%5E%7B2%7Ds)
Also at 1500 K,
= ![2.82 \times 10^{5} \times e^({\frac{-9835}{1500}}) m^{6}/mol^{2}s](https://tex.z-dn.net/?f=2.82%20%5Ctimes%2010%5E%7B5%7D%20%5Ctimes%20e%5E%28%7B%5Cfrac%7B-9835%7D%7B1500%7D%7D%29%20m%5E%7B6%7D%2Fmol%5E%7B2%7Ds)
![K_{r} = 400.613 m^{6}/mol^{2}s](https://tex.z-dn.net/?f=K_%7Br%7D%20%3D%20400.613%20m%5E%7B6%7D%2Fmol%5E%7B2%7Ds)
Relation between
and
is as follows.
![K_{p} = K_{c}RT](https://tex.z-dn.net/?f=K_%7Bp%7D%20%3D%20K_%7Bc%7DRT)
Putting the given values into the above formula as follows.
![K_{p} = K_{c}RT](https://tex.z-dn.net/?f=K_%7Bp%7D%20%3D%20K_%7Bc%7DRT)
![0.003691 = K_{c} \times 8.314 \times 1500](https://tex.z-dn.net/?f=0.003691%20%3D%20K_%7Bc%7D%20%5Ctimes%208.314%20%5Ctimes%201500)
![K_{c} = 2.9 \times 10^{-7}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%202.9%20%5Ctimes%2010%5E%7B-7%7D)
Also, ![K_{c} = \frac{K_{f}}{K_{r}}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%20%5Cfrac%7BK_%7Bf%7D%7D%7BK_%7Br%7D%7D)
or, ![K_{f} = K_{c} \times K_{r}](https://tex.z-dn.net/?f=K_%7Bf%7D%20%3D%20K_%7Bc%7D%20%5Ctimes%20K_%7Br%7D)
= ![2.9 \times 10^{-7} \times 400.613](https://tex.z-dn.net/?f=2.9%20%5Ctimes%2010%5E%7B-7%7D%20%5Ctimes%20400.613)
= ![1.1617 \times 10^{-4} m^{6}/mol^{2}s](https://tex.z-dn.net/?f=1.1617%20%5Ctimes%2010%5E%7B-4%7D%20m%5E%7B6%7D%2Fmol%5E%7B2%7Ds)
Thus, we can conclude that the value of
is
.
Maybe if you put it in english i’ll answer.. nahh jp i’m fluent
Answer:
V₂ = 2.96 L
Explanation:
Given data:
Initial volume = 2.00 L
Initial temperature = 250°C
Final volume = ?
Final temperature = 500°C
Solution:
First of all we will convert the temperature into kelvin.
250+273 = 523 k
500+273= 773 k
According to Charles's law,
V∝ T
V = KT
V₁/T₁ = V₂/T₂
V₂ = T₂V₁/T₁
V₂ = 2 L × 773 K / 523 k
V₂ = 1546 L.K / 523 k
V₂ = 2.96 L