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Artist 52 [7]
3 years ago
13

When a quantity of monatomic ideal gas expands at a constant pressure of 4.00×104pa, the volume of the gas increases from 2.00×1

0−3m3 to 8.00×10−3m3?
Physics
2 answers:
3241004551 [841]3 years ago
7 0
This can be verified if we know the values of the initial (T1) and final (T2) temperatures. We use the ideal gas equation for this: PV=RT.

P1V1=RT1
(40000 Pa)*(0.002 m^3) = (8.314 m3Pa/molK)(T1)
T1 = 9.62 K


P2V2=RT2
(40000 Pa)*(0.008 m^3) = (8.314 m3Pa/molK)(T2)
T2 = 33.5 K

Thus, this is true if the monoatomic ideal gas is heated from 9.62 K to 33.5 K at constant pressure.
Semmy [17]3 years ago
5 0
Yes that is correct. We know this because 4.00 x 10 4 Pa is constant. If you have 2.00×10−3m3 then you do the following: (2.00×10^−3)(4.00×10^<span> 4) = </span>8.00×10^−3. That is how you get your answer
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7 0
3 years ago
Two charged particles, with charges q1=q and q2=4q, are located at a distance d= 2.00cm apart on the x axis. A third charged par
Murrr4er [49]

Answer:

X₃₁ = 0.58 m  and  X₃₂ = -1.38 m

Explanation:

For this exercise we use Newton's second law where the force is the Coulomb force

        F₁₃ - F₂₃ = 0

        F₁₃ = F₂₃

Since all charges are of the same sign, forces are repulsive

        F₁₃ = k q₁ q₃ / r₁₃²

        F₂₃ = k q₂ q₃ / r₂₃²

Let's find the distances

         r₁₃ = x₃- 0

         r₂₃ = 2 –x₃

We substitute

      k q q / x₃² = k 4q q / (2-x₃)²

      q² (2 - x₃)² = 4 q² x₃²

        4- 4x₃ + x₃² = 4 x₃²

        5x₃² + 4 x₃ - 4 = 0

We solve the quadratic equation

        x₃ = [-4 ±√(16 - 4 5 (-4)) ] / 2  5

        x₃ = [-4 ± 9.80] 10

       X₃₁ = 0.58 m

       X₃₂ = -1.38 m

For this two distance it is given that the two forces are equal

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3 years ago
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