Answer:
A) 1059 J/mol
B) 17,920 J/mol
Explanation:
Given that:
Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4
R (constant) = 8.314
We know that:
![C_p=C_v+R](https://tex.z-dn.net/?f=C_p%3DC_v%2BR)
We can determine
from above if we make
the subject of the formula as:
![C_v](https://tex.z-dn.net/?f=C_v)
![=C_p-R](https://tex.z-dn.net/?f=%3DC_p-R)
![C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314](https://tex.z-dn.net/?f=C_V%20%3D%2029.42-%282.7%2A10%5E%7B-3%7D%29T%2B%285.82%2A10%5E%7B-7%7D%29T2-%281.305%2A10%5E%7B-8%7D%29T3-%288.23%2A10%5E%7B-12%7D%29T4-8.314)
![C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4](https://tex.z-dn.net/?f=C_V%20%3D%2021.106-%282.7%2A10%5E%7B-3%7D%29T%2B%285.82%2A10%5E%7B-7%7D%29T2-%281.305%2A10%5E%7B-8%7D%29T3-%288.23%2A10%5E%7B-12%7D%29T4)
A).
The formula for calculating change in internal energy is given as:
![dU=C_vdT](https://tex.z-dn.net/?f=dU%3DC_vdT)
If we integrate above data into the equation; it implies that:
![U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du](https://tex.z-dn.net/?f=U2-U1%3D%5Cint%5Climits%5E%7B500%7D_%7B450%7D%2821.106-%282.7%2A10%5E%7B-3%7D%29T%2B%285.82%2A10%5E%7B-7%7D%29T2-%281.305%2A10%5E%7B-8%7D%29T3-%288.23%2A10%5E%7B-12%7D%29T4%5C%2C%29%20du)
![U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)](https://tex.z-dn.net/?f=U2-U1%3D%5Cint%5Climits%5E%7B500%7D_%7B450%7D%2821.106-%282.7%2A10%5E%7B-3%7D%29T%2F1%2B%285.82%2A10%5E%7B-7%7D%29T2%2F2-%281.305%2A10%5E%7B-8%7D%29T3%2F3-%288.23%2A10%5E%7B-12%7D%29T4%2F4%5C%2C%29)
![U2-U1= 1059J/mol](https://tex.z-dn.net/?f=U2-U1%3D%201059J%2Fmol)
Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.
B).
If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.
then T = 273 K & T2 = 1073 K
∴
![U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)](https://tex.z-dn.net/?f=U2-U1%3D%5Cint%5Climits%5E%7B500%7D_%7B450%7D%2821.106-%282.7%2A10%5E%7B-3%7D%29T%2F1%2B%285.82%2A10%5E%7B-7%7D%29T2%2F2-%281.305%2A10%5E%7B-8%7D%29T3%2F3-%288.23%2A10%5E%7B-12%7D%29T4%2F4%5C%2C%29)
![U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)](https://tex.z-dn.net/?f=U2-U1%3D%5Cint%5Climits%5E%7B500%7D_%7B450%7D%2821.106-%282.7%2A10%5E%7B-3%7D%29273%2F1%2B%285.82%2A10%5E%7B-7%7D%291073%2F2-%281.305%2A10%5E%7B-8%7D%29T3%2F3-%288.23%2A10%5E%7B-12%7D%29T4%2F4%5C%2C%29)
![U2-U1= 17,920 J/mol](https://tex.z-dn.net/?f=U2-U1%3D%2017%2C920%20J%2Fmol)
In your question where the ask is to compute the specific heat capacity (in J/Kg*K) at the constant volume of nitrogen(n_2) gas and the molar mass of N_2 is 28.014 g/mol and in my calculation i came up with an answer of 890.391 J/Kg*K
Answer:
0.006342moles
Explanation:
1000ml of NaOH contain 0.151moles
42ml of NaOH contain (42*0.151)/1000 moles
=0.006342moles