Answer:
The vapor pressure at 60.6°C is 330.89 mmHg
Explanation:
Applying Clausius Clapeyron Equation
![ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}]](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%20%3D%20%5Cfrac%7B%5Cdelta%20H%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%20%5Cfrac%7B1%7D%7BT_2%7D%5D)
Where;
P₂ is the final vapor pressure of benzene = ?
P₁ is the initial vapor pressure of benzene = 40.1 mmHg
T₂ is the final temperature of benzene = 60.6°C = 333.6 K
T₁ is the initial temperature of benzene = 7.6°C = 280.6 K
ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol
R is gas rate = 8.314 J/mol.k
![ln(\frac{P_2}{40.1}) = \frac{31,000}{8.314}[\frac{1}{280.6}- \frac{1}{333.6}]\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.003564 - 0.002998)\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.000566)\\\\ln(\frac{P_2}{40.1}) = 2.1104\\\\\frac{P_2}{40.1} = e^{2.1104}\\\\\frac{P_2}{40.1} = 8.2515\\\\P_2 = (40.1*8.2515)mmHg = 330.89 mmHg](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%20%5Cfrac%7B31%2C000%7D%7B8.314%7D%5B%5Cfrac%7B1%7D%7B280.6%7D-%20%5Cfrac%7B1%7D%7B333.6%7D%5D%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%203728.65%20%280.003564%20-%200.002998%29%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%203728.65%20%20%280.000566%29%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%202.1104%5C%5C%5C%5C%5Cfrac%7BP_2%7D%7B40.1%7D%20%3D%20e%5E%7B2.1104%7D%5C%5C%5C%5C%5Cfrac%7BP_2%7D%7B40.1%7D%20%3D%208.2515%5C%5C%5C%5CP_2%20%3D%20%2840.1%2A8.2515%29mmHg%20%3D%20330.89%20mmHg)
Therefore, the vapor pressure at 60.6°C is 330.89 mmHg
Answer:
2,54 cm are equal to 1 inch
Explanation:
Doing the conversion:
![55[cm]*\frac{1[inch]}{2,54[cm]} =21,65[inch]](https://tex.z-dn.net/?f=55%5Bcm%5D%2A%5Cfrac%7B1%5Binch%5D%7D%7B2%2C54%5Bcm%5D%7D%20%3D21%2C65%5Binch%5D)
Explanation:
Formula to represent thrust is as follows.
F = ![\frac{dP}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7BdP%7D%7Bdt%7D)
= ![\frac{2p}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B2p%7D%7Bdt%7D)
or, p = ![\frac{E}{c}](https://tex.z-dn.net/?f=%5Cfrac%7BE%7D%7Bc%7D)
![\frac{p}{dt} = \frac{W}{c}](https://tex.z-dn.net/?f=%5Cfrac%7Bp%7D%7Bdt%7D%20%3D%20%5Cfrac%7BW%7D%7Bc%7D)
F = ![\frac{2IA}{c}](https://tex.z-dn.net/?f=%5Cfrac%7B2IA%7D%7Bc%7D)
= ![\frac{2 (1000 W/m^{2})(5.5 \times 10^{3} m)^{2}}{3 \times 10^{8} m/s}](https://tex.z-dn.net/?f=%5Cfrac%7B2%20%281000%20W%2Fm%5E%7B2%7D%29%285.5%20%5Ctimes%2010%5E%7B3%7D%20m%29%5E%7B2%7D%7D%7B3%20%5Ctimes%2010%5E%7B8%7D%20m%2Fs%7D)
= 201.67 N
Thus, we can conclude that the thrust is 201.67 N.
The bed load moves the slowest from all the parts of the stream's sediment. It consists of particles suspended that are suspended and float around the bed. This part is the slowest in motion, as it rolls, and moves with the flow. The particles near the bed are not dissolved so they settle at the bottom and move with the stream.