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snow_lady [41]
4 years ago
9

Consider an ideal gas of 7 moles that is in contact with a thermal reservoir of temperature 475 K. The gas is enclosed in a cont

ainer that is allowed to expand and contract. If the initial volume of the gas is 0.50 m ³, find the pressure of the gas some time later after the volume has expanded to 1.50 m ³.
Physics
1 answer:
kozerog [31]4 years ago
3 0

Answer:

(a). The initial pressure is 5.5\times10^{4}\ Pa

(b). The final pressure is 1.8\times10^{4}\ Pa

Explanation:

Given that,

Number of moles = 7

Temperature = 475 K

Initial volume = 0.50 m³

Expanded volume = 1.50 m³

We need to calculate the initial pressure

Using formula of pressure

P_{i}=\dfrac{nRT_{i}}{V_{i}}

Put the value into the formula

P_{i}=\dfrac{7\times8.31\times475}{0.50}

P_{i}=55261.5\ Pa

P_{i}=5.5\times10^{4}\ Pa

We need to calculate the final pressure

Using formula of pressure

P_{f}V_{f}=nRT_{f}

After expansion,

\dfrac{P_{f}V_{f}}{P_{i}V_{i}}=\dfrac{nRT_{f}}{nRT_{i}}

P_{f}=\dfrac{T_{f}}{T_{i}}\times\dfrac{P_{i}V_{i}}{V_{f}}

Put the value into the formula

For thermal process,

T_{i}=T_{f}

P_{f}=\dfrac{5.5\times10^{4}\times0.50}{1.50}

P_{f}=18333.33\ Pa

P_{f}=1.8\times10^{4}\ Pa

Hence, (a). The initial pressure is 5.5\times10^{4}\ Pa

(b). The final pressure is 1.8\times10^{4}\ Pa

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Answer:

Answer:

A. - 0.017N. It acts to the left.

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Explanation:

A. For the +65μC charge, we consider it to be the origin. Hence, the two other charges are on the +x axis.

The net coulombs force on the charge is

F = [KQ(1)Q(2)]/(r^2) + [KQ(1)Q(3)]/(r^2)

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Q(3) = charge on the rightmost side.

F = [(8.988 × 10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(-95×10^-6)×(65×10^-6)]/(40^2)

F = 0.01753 - 0.03469

F = -0.017N

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B. For the +48μC charge, we consider it to be the origin. Hence, the leftmost charge is on the - x axis and the rightmost charge is on the +x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) + [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(48×10^-6)]/(40^2) + [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = -0.017 - 0.02562

F = - 0.043N

It has a negative sign, hence, it acts to the left.

C. For the -95μC charge, we consider it to be the origin. Hence, the two other charges are on the - x axis.

The net coulombs force on the charge is

F = [-KQ(1)Q(3)]/(r^2) - [KQ(2)Q(3)]/(r^2)

F = [-(8.988×10^9)×(65×10^-6)×(-95×10^-6)]/(40^2) - [(8.988 × 10^9)×(48×10^-6)×(-95×10^-6)]/(40^2)

F = +0.03469 + 0.02562

F = +0.060N

It has a positive sign, hence, it acts to the right.

Read more on Brainly.com - brainly.com/question/14592748#readmore

Explanation:

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By definition of energy efficiency, we derive an expression for the energy rate exhausted to the river (Q_{out}), in megawatts:

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Q_{out} = \left(\frac{1}{\eta}-1 \right)\cdot W(1)

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If we know that \eta = 0.39 and W = 330\,MW, then the energy rate exhausted to the river is:

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Q_{out} = 516.154\,MW

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