Ask question

Ask question

All categories

3 years ago

14

The vapor pressure of ethanol at 293 K is 5.95 kPa and at 336.5 K it is 53.3 kPa. Calculate the enthalpy of vaporization of etha

nol. (Note: any unit of pressure may be used, as long as you are doing P2/P1. Temp must be K)

1 answer:

denis-greek [22]3 years ago

4 0

Answer:

$H=41.3kJmol^{-1}$

Explanation:

The equation relating the the enthalphy, pressure and temperature is expressed as

$ln(\frac{P_{2}}{P_{1}} )=\frac{H}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}} ) \\$

Where P is the pressure, H is the enthalphy, and T is the temperature.

since the given values are

$T_{1}=293k, \\T_{2}=336.5k\\P_{1}=5.95kPA\\p_{2}=53.3kPA\\and R=8.314J.K^{-1}mol_{-1}$

if we insert values, we arrive at

$ln(\frac{53.3}{5.95} )=\frac{H}{8.314}(\frac{1}{293}-\frac{1}{336.5} )\\2.19=\frac{H}{8.314}(0.00044)\\H=(2.19*8.314)/0.00044\\H=41,268.8Jmol^{-1}\\H=41.3kJmol^{-1}$

You might be interested in

Why are there temperature differences on the moon's surface even though there is no atmosphere present?

nika2105 [10]

The lack of an atmosphere means convection cannot happen on the moon. Therefore, there is no form of heat dissipation on regions in direct sunlight. In addition, the lack of an atmosphere means there is no greenhouse effect on the moon. This is why regions facing away from sunlight are very cold.

4 0

3 years ago

I need help finding the answer

miss Akunina [59]

Speed = (distance) / (time)

Speed = (2.3 m) / (3 sec)

Speed = (2.3/3) (m/s)

<em>Speed = 0.766... m/s</em>

5 0

3 years ago

Two point charges of +2.50 x 10^-5 C and -2.50 x 10^-5 C are separated by 0.50m. Which of the following describes the force betw

mr Goodwill [35]

Answer:

Q1 = +2.50 x 10^-5C and Q2 = -2.50 x 10^-5C, r = 0.50m, F=?

Using Coulomb's law:

F = 1/(4πE) x Q1 x Q2/ r^2

Where

k= 1/(4πE) = 9 x 10^9Nm2/C2

Therefore,

F = 9x 10^9 x 2.50 x 10^-5 x2.50 x

10^-5/. ( 0.5)^2

F= 5.625/ 0.25

F= 22.5N approximately

F= 23N.

To find the direction of the force: since Q1 is positive and Q2 is negative, the force along Q1 and Q2 is force of attraction.

Hence To = 23N, attractive. C ans.

Thanks.

5 0

3 years ago

Determine the magnitude and direction of the resultant force of the following free body diagram.

Papessa [141]

Answer:

The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.

Explanation:

First, we must calculate the resultant force ( $\vec F$ ), in newtons, by vectorial sum:

$\vec F = [(-200\,N)\cdot \cos 60^{\circ}+(400\,N)\cdot \cos 45^{\circ}+300\,N]\,\hat{i} + [(200\,N)\cdot \sin 60^{\circ} + (400\,N)\cdot \sin 45^{\circ}-100\,N]\,\hat{j}$ (1)

$\vec F = 182.843\,\hat{i} + 356.048\,\hat{j}$

Second, we calculate the magnitude of the resultant force by Pythagorean Theorem:

$\|\vec F\| = \sqrt{(482.843\,N)^{2}+(356.048\,N)^{2}}$

$\|\vec F\| \approx 599.923\,N$

Let suppose that direction of the resultant force is an standard angle. According to (1), the resultant force is set in the first quadrant:

$\theta = \tan^{-1}\left(\frac{356.048\,N}{482.843\,N} \right)$

Where $\theta$ is the direction of the resultant force, in sexagesimal degrees.

$\theta \approx 36.405^{\circ}$

The magnitude and direction of the resultant force are approximately 599.923 newtons and 36.405°.

4 0

3 years ago

A reversible Carnot engine has an efficiency of 30% and it operates between a heat source maintained at T 1 , and a refrigerator

bixtya [17]

To solve this problem we will apply the concepts related to the thermal efficiency given in an engine of the Carnot cycle. Here we know that efficiency is given under the equation

$\eta = 1 - \frac{T_2}{T_1}$

Where,

$T_2 =$ Temperature of Cold Body

$T_1 =$ Temperature of Hot Body

$\eta$ = Efficiency

According to the statement our values are:

$\eta = 0.3$

$T_2 = 210K$

Replacing we have that

$0.3 = 1- \frac{210}{T_1}$

$\frac{210}{T_1} = 0.7$

$T_1 = \frac{210}{0.7}$

$T_1 = 300K$

Therefore the temperature of the heat source is 300K

6 0

3 years ago