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3 years ago

9

A quantity of an ideal gas is kept in a rigid container of constant volume. If the gas is originally at a temperature of 28 °C,

at what temperature (in °C) will the pressure of the gas triple from its base value?

1 answer:

Gnoma [55]3 years ago

8 0

Answer:

$T_2=630^{\circ}C$ '

Explanation:

Original temperature of the gas, $T_1=28^{\circ}C=301\ K$

From the ideal gas equation,

$P_1V_1=nRT_1$

Since,

$P_2=3P_1$

$nRT_2=3(nRT_1)$

$T_2=3T_1$

$T_2=3\times 301$

$T_2=903\ K$

or

$T_2=630^{\circ}C$

So, the new temperature of the gas is 630 degree Celsius. Hence, this is the required solution.

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Determine the minimum work per unit of heat transfer from the source reservoir that is required to drive a heat pump with therma

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Answer:

The minimum work per unit heat transfer will be 0.15.

Explanation:

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3 years ago

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.57 A out of the jun

AlekseyPX

Answer:

a. 1.56 × 10¹⁸ electrons per second

b. The electrons in wire 3 flow into the junction.

Explanation:

Here is the complete question

Three wires meet at a junction. Wire 1 has a current of 0.40 A into the junction. The current of wire 2 is 0.65 A out of the junction. (a) How many electrons per second move past a point in wire 3? (b) In which direction do the electrons move in wire 3 -- into or out of the junction?

Solution

(a) How many electrons per second move past a point in wire 3?

Using Kirchhoff's current law, at the junction, i₁ + i₂ + i₃ = 0 where i₁ = current in wire 1 = 0.40 A, i₂ = current in wire 2 = 0.65 A and i₃ = = current in wire 3,

So, i₃ = -(i₁ + i₂)

taking current flowing into the junction as positive and those leaving as negative, i₁ = + 0.40 A and i₂ = -0.65 A

So, i₃ = -(i₁ + i₂)

i₃ = -(0.40 A + (-0.65 A))

i₃ = -(0.40 A - 0.65 A)

i₃ = -(-0.25 A)

i₃ = 0.25 A

Since i₃ = 0.25 C/s and we have e = 1.602 × 10⁻¹⁹ C per electron, then the number of electrons flowing in wire 3 per second is i₃/e = 0.25 C/s ÷ 1.602 × 10⁻¹⁹ C per electron = 0.1561 × 10¹⁹ electrons per second = 1.561 × 10¹⁸ electrons per second ≅ 1.56 × 10¹⁸ electrons per second

(b) In which direction do the electrons move -- into or out of the junction?

Given that i₃ = + 0.25 A and that positive flows into the junction, thus, the electrons in wire 3 flow into the junction.

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