Answer: The zeroth law of thermodynamics states that if the two thermodynamic systems are in equilibrium with a third system, then they are also in thermal equilibrium with each other.
Explanation: The third law of thermodynamics is basically used for the measurement of temperature. It states that if the two objects A and B, are in equilibrium with the third object C, then A and B objects are in equilibrium with each other. Basically, it would mean that all three A, B and C are at the same temperature. It is also used to compare the temperature of multiple objects.
Sounds to me that it’s most likely locked up and you need a new one
Answer:
when you are in the design process so you can get his insite
Explanation:
Answer:
a)W=12.62 kJ/mol
b)W=12.59 kJ/mol
Explanation:
At T = 100 °C the second and third virial coefficients are
B = -242.5 cm^3 mol^-1
C = 25200 cm^6 mo1^-2
Now according isothermal work of one mole methyl gas is
W=-![\int\limits^a_b {P} \, dV](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7BP%7D%20%5C%2C%20dV)
a=![v_2\\](https://tex.z-dn.net/?f=v_2%5C%5C)
b=![v_1](https://tex.z-dn.net/?f=v_1)
from virial equation
![\frac{PV}{RT}=z=1+\frac{B}{V}+\frac{C}{V^2}\\ \\P=RT(1+\frac{B}{V} +\frac{C}{V^2})\frac{1}{V}\\](https://tex.z-dn.net/?f=%5Cfrac%7BPV%7D%7BRT%7D%3Dz%3D1%2B%5Cfrac%7BB%7D%7BV%7D%2B%5Cfrac%7BC%7D%7BV%5E2%7D%5C%5C%20%20%20%5C%5CP%3DRT%281%2B%5Cfrac%7BB%7D%7BV%7D%20%2B%5Cfrac%7BC%7D%7BV%5E2%7D%29%5Cfrac%7B1%7D%7BV%7D%5C%5C)
And
![W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV](https://tex.z-dn.net/?f=W%3D-%5Cint%5Climits%5Ea_b%20%7BRT%281%2B%5Cfrac%7BB%7D%7BV%7D%20%2B%5Cfrac%7BC%7D%7BV%5E2%7D%5Cfrac%7B1%7D%7BV%7D%20%20%7D%20%5C%2C%20dV)
a=![v_2\\](https://tex.z-dn.net/?f=v_2%5C%5C)
b=![v_1](https://tex.z-dn.net/?f=v_1)
Now calculate V1 and V2 at given condition
![\frac{P1V1}{RT} = 1+\frac{B}{v_1} +\frac{C}{v_1^2}](https://tex.z-dn.net/?f=%5Cfrac%7BP1V1%7D%7BRT%7D%20%3D%201%2B%5Cfrac%7BB%7D%7Bv_1%7D%20%2B%5Cfrac%7BC%7D%7Bv_1%5E2%7D)
Substitute given values
= 1 x 10^5 , T = 373.15 and given values of coefficients we get
![10^5(v_1)/8.314*373.15=1-242.5/v_1+25200/v_1^2](https://tex.z-dn.net/?f=10%5E5%28v_1%29%2F8.314%2A373.15%3D1-242.5%2Fv_1%2B25200%2Fv_1%5E2)
Solve for V1 by iterative or alternative cubic equation solver we get
![v_1=30780 cm^3/mol](https://tex.z-dn.net/?f=v_1%3D30780%20cm%5E3%2Fmol)
Similarly solve for state 2 at P2 = 50 bar we get
![v_1=241.33 cm^3/mol](https://tex.z-dn.net/?f=v_1%3D241.33%20cm%5E3%2Fmol)
Now
![W=-\int\limits^a_b {RT(1+\frac{B}{V} +\frac{C}{V^2}\frac{1}{V} } \, dV](https://tex.z-dn.net/?f=W%3D-%5Cint%5Climits%5Ea_b%20%7BRT%281%2B%5Cfrac%7BB%7D%7BV%7D%20%2B%5Cfrac%7BC%7D%7BV%5E2%7D%5Cfrac%7B1%7D%7BV%7D%20%20%7D%20%5C%2C%20dV)
a=241.33
b=30780
After performing integration we get work done on the system is
W=12.62 kJ/mol
(b) for Z = 1 + B' P +C' P^2 = PV/RT by performing differential we get
dV=RT(-1/p^2+0+C')dP
Hence work done on the system is
![W=-\int\limits^a_b {P(RT(-1/p^2+0+C')} \, dP](https://tex.z-dn.net/?f=W%3D-%5Cint%5Climits%5Ea_b%20%7BP%28RT%28-1%2Fp%5E2%2B0%2BC%27%29%7D%20%5C%2C%20dP)
a=![v_2\\](https://tex.z-dn.net/?f=v_2%5C%5C)
b=![v_1](https://tex.z-dn.net/?f=v_1)
by substituting given limit and P = 1 bar , P2 = 50 bar and T = 373 K we get work
W=12.59 kJ/mol
The work by differ between a and b because the conversion of constant of virial coefficients are valid only for infinite series
Answer:a government website/A.Gov
Explanation:
I looked it up I also put this as my answer so i don’t know if it’s right or not