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Softa [21]
3 years ago
5

A cylinder with a piston contains 0.250 mol of oxygen at2.40x105 Pa and 355 K. The oxygen may be treated as an ideal gas.The gas

first expands isobarically to twice its original volume. Itis then compressed isothermally back to its original volume, andfinally it is cooled isochorically to its original pressure.
(a)Show the series of processes on a pV diagram.
(b) Compute thetemperature during the isothermal compression.
(c) Compute themaximum pressure.
(d) Compute the total work done by the piston onthe gas during the series of processes.

Physics
1 answer:
Nikitich [7]3 years ago
7 0

Answer:

= 285 Joules

Explanation:

a) answer can be found out in attachment

(b) The temperature for the isothermal compression is the same as the temp at the end of the isobaric expansion. Since pressure is held constant but volume doubles, we use the ideal gas law:

p V = nR T          to see that the temperature also doubles.

.So...   temp for isothermal compression =   355×2 = 710 K

.(c)   The max pressure occurs at the top point. At this point, the volume is back to the original value but the temperature is twice the original value. So the pressure at this point is twice the original, or

max pressure = 2×240000 Pa =  480000 Pa  =   4.80 x 10^5 Pa

(d) total work done by the piston = workdone during isothermal compression - work done during expansion =

=  nRT ln(V initial / V final)-p (V initial - V final)

=   nRT ln(2) - nR(T final - T initial)  

= 0.250× 8.314 ×710×ln(2)-0.250×8.314× (710 - 355)

=     285 Joules

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Coulomb’s law and static point charge ensembles (15 points). A test charge of 2C is located at point (3, 3, 5) in Cartesian coor
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Answer:

a) F_{r}= -583.72MN i + 183.47MN j + 6.05GN k

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Step-by-step explanation.

In order to solve this problem, we mus start by plotting the given points and charges. That will help us visualize the problem better and determine the direction of the forces (see attached picture).

Once we drew the points, we can start calculating the forces:

r_{AP}^{2}=(3-0)^{2}+(3-0)^{2}+(5+0)^{2}

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Next, we can make use of the force formula:

F=k_{e}\frac{q_{1}q_{2}}{r^{2}}

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F_{AP}=(8.99x10^{9})\frac{(1C)(2C)}{43m^{2}}

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F_{APx}=418.14 MN*\frac{3}{\sqrt{43}}i

F_{APx}=191.30 MNi

F_{APy}=418.14 MN*\frac{3}{\sqrt{43}}j

F_{APy}=191.30MN j

F_{APz}=418.14 MN*\frac{5}{\sqrt{43}}k

F_{APz}=318.83 MN k

And we can now write them together for the first force, so we get:

F_{AP}=(191.30i+191.30j+318.83k)MN

We continue with the next force. The procedure is the same so we get:

r_{BP}^{2}=(3-1)^{2}+(3-1)^{2}+(5+0)^{2}

which yields:

r_{BP}^{2}= 33 m^{2}

Next, we can make use of the force formula:

F_{BP}=(8.99x10^{9})\frac{(4C)(2C)}{33m^{2}}

which yields:

F_{BP}=2.18 GN

Now we can find its components:

F_{BPx}=2.18 GN*\frac{2}{\sqrt{33}}i

F_{BPx}=758.98 MNi

F_{BPy}=2.18 GN*\frac{2}{\sqrt{33}}j

F_{BPy}=758.98MN j

F_{BPz}=2.18 GN*\frac{5}{\sqrt{33}}k

F_{BPz}=1.897 GN k

And we can now write them together for the second, so we get:

F_{BP}=(758.98i + 758.98j + 1897k)MN

We continue with the next force. The procedure is the same so we get:

r_{CP}^{2}=(3-5)^{2}+(3-4)^{2}+(5-0)^{2}

which yields:

r_{CP}^{2}= 30 m^{2}

Next, we can make use of the force formula:

F_{CP}=(8.99x10^{9})\frac{(7C)(2C)}{30m^{2}}

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F_{CP}=4.20 GN

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F_{CPx}=4.20 GN*\frac{-2}{\sqrt{30}}i

F_{CPx}=-1.534 GNi

F_{CPy}=4.20 GN*\frac{2}{\sqrt{30}}j

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F_{CPz}=4.20 GN*\frac{5}{\sqrt{30}}k

F_{CPz}=3.83 GN k

And we can now write them together for the third force, so we get:

F_{CP}=(-1.534i - 0.76681j +3.83k)GN

So in order to find the resultant force, we need to add the forces together:

F_{r}=F_{AP}+F_{BP}+F_{CP}

so we get:

F_{r}=(191.30i+191.30j+318.83k)MN + (758.98i + 758.98j + 1897k)MN + (-1.534i - 0.76681j +3.83k)GN

So when adding the problem together we get that:

F_{r}=(-0.583.72i + 0.18347j +6.05k)GN

which is the answer to part a), now let's take a look at part b).

b)

Basically, we need to find the magnitude of the force and divide it into the test charge, so we get:

F_{r}=\sqrt{(-0.583.72)^{2} + (0.18347)^{2} +(6.05)^{2}}

which yields:

F_{r}=6.08 GN

and now we take the formula for the electric field which is:

E=\frac{F_{r}}{q}

so we go ahead and substitute:

E=\frac{6.08GN}{2C}

E=3.04\frac{GN}{C}

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