Answer:
C. 26.4 kJ/mol
Explanation:
The Chen's rule for the calculation of heat of vaporization is shown below:
![\Delta H_v=RT_b\left [ \frac{3.974\left ( \frac{T_b}{T_c} \right )-3.958+1.555lnP_c}{1.07-\left ( \frac{T_b}{T_c} \right )} \right ]](https://tex.z-dn.net/?f=%5CDelta%20H_v%3DRT_b%5Cleft%20%5B%20%5Cfrac%7B3.974%5Cleft%20%28%20%5Cfrac%7BT_b%7D%7BT_c%7D%20%5Cright%20%29-3.958%2B1.555lnP_c%7D%7B1.07-%5Cleft%20%28%20%5Cfrac%7BT_b%7D%7BT_c%7D%20%5Cright%20%29%7D%20%5Cright%20%5D)
Where,
is the Heat of vaoprization (J/mol)
is the normal boiling point of the gas (K)
is the Critical temperature of the gas (K)
is the Critical pressure of the gas (bar)
R is the gas constant (8.314 J/Kmol)
For diethyl ether:



Applying the above equation to find heat of vaporization as:
![\Delta H_v=8.314\times307.4 \left [ \frac{3.974\left ( \frac{307.4}{466.7} \right )-3.958+1.555ln36.4}{1.07-\left ( \frac{307.4}{466.7} \right )} \right ]](https://tex.z-dn.net/?f=%5CDelta%20H_v%3D8.314%5Ctimes307.4%20%5Cleft%20%5B%20%5Cfrac%7B3.974%5Cleft%20%28%20%5Cfrac%7B307.4%7D%7B466.7%7D%20%5Cright%20%29-3.958%2B1.555ln36.4%7D%7B1.07-%5Cleft%20%28%20%5Cfrac%7B307.4%7D%7B466.7%7D%20%5Cright%20%29%7D%20%5Cright%20%5D)

The conversion of J into kJ is shown below:
1 J = 10⁻³ kJ
Thus,

<u>Option C is correct</u>
Answer:- New pressure is 0.942 atm.
Solution:- The volume of the glass bottle would remain constant here and the pressure will change with the temperature.
Pressure is directly proportional to the kelvin temperature. The equation used here is:

Where,
and
are initial and final temperatures,
and
are initial and final pressures.
= 20.3 + 273.15 = 293.45 K
= -2.0 + 273.15 = 271.15 K
= 1.02 atm
= ?
Let's plug in the values in the equation and solve it for final pressure.


= 0.942 atm
So, the new pressure of the jar is 0.942 atm.
Answer:
Well they help us understand the properties of matter of course!