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andrey2020 [161]

2 years ago

10

Consider an insulated tank with a volume V = 2 L is separated into two equal-volume parts by a thin wall. On the left is an idea

l gas at P = 100kPa and T = 500K; on the right is a vacuum. Now the wall is removed. a) What is the final temperature in the tank? b) What is the ∆Suniv due to the removal of the wall?

1 answer:

steposvetlana [31]2 years ago

5 0

Answer

given,

V = 2 L

the left is an ideal gas at P = 100 k Pa and T = 500 K

mass is constant

$m_1 = m_2$

$\dfrac{P_1V_1}{RT_1} = \dfrac{P_2V_2}{RT_2}$

Pressure is same because it's not changing due to process

$\dfrac{V}{500} = \dfrac{2 V}{T_2}$

$T_2 = 1000\ K$

$\Delta S_{univ} = \Delta S_{sys} + (\Delta S)_{surr}$

$\Delta S_{univ} =m(C_v ln (\dfrac{T_2}{T_1}))+ R ln (\dfrac{V_2}{V_1})$

$m = \dfrac{P_1V_1}{RT_1}$

$m = \dfrac{100 \times 10^3 \times 2 \times 10^{-3}}{287\times 500}$

m = 1.39 x 10⁻³ Kg

$\Delta S_{univ} =1.39\times 10^{-3}(0.718 ln\ 2+ 0.287 ln (2)$

$\Delta S_{univ} =0.968\times 10^{-3}\ kJ/K$

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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:

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which yields:

$r_{AP}^{2}= 43 m^{2}$

(I will assume the positions are in meters)

Next, we can make use of the force formula:

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Now we can find its components:

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

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

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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:

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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:

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

$F_{BPx}=758.98 MNi$

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

$F_{BPy}=758.98MN j$

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

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

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

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which yields:

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Next, we can make use of the force formula:

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which yields:

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

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

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$F_{CPz}=3.83 GN k$

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

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

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

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

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

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