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Debora [2.8K]
3 years ago
10

A 1.5-mole sample of an ideal gas is gently cooled at constant temperature 320 K. It contracts

Physics
1 answer:
Alona [7]3 years ago
6 0

Answer:

V₂ = 14.07 L

Explanation:

As this gas is cooled at constant temperature of 320 K, this means that we are on an isothermal process, and according to the 1st law of thermodynamics:

Q = W   (1)

And as the temperature is constant, we can use the following expression to calculate the Work done:

W = nRT ln(V₁/V₂)   (2)

However, as Q = W, we can replace heat into the above expression and then solve for V₂:

Q = nRT ln(V₁/V₂)

Replacing we have:

1200 = (1.5 * 8.314 * 320) ln(19/V₂)

1200 = 39907.2 ln(19/V₂)

ln(19/V₂) = 1200/3990.72

ln(19/V₂) = 0.3007

19/V₂ = e^(0.3007)

V₂ = 19 / e^(0.3007)

<h2>V₂ = 14.07 L</h2>

Hope this helps

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M=3000km v=25m/s what’s the momentum
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<h3>Answer:</h3>

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<h3>Explanation:</h3>

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A particle with charge −− 5.90 nCnC is moving in a uniform magnetic field B⃗ =−(B→=−( 1.28 TT )k^k^. The magnetic force on the p
maks197457 [2]

Answer:

Explanation:

Given that,

Charge q=-5.90nC

Magnetic field B= -1.28T k

And the magnetic force

F =−( 3.70×10−7N )i+( 7.60×10−7N )j

Let the velocity be V(xi + yj + zk)

Then, the force is given as

Note i×i=j×j×k×k=0

i×j=k. j×i=-k

j×k=i. k×j=-i

k×i=j. i×k=-j

The force in a magnetic field is given as

F= q(v×B)

−( 3.70×10−7N )i+( 7.60×10−7N )j =

q(xi + yj + zk) × -1.28k

−( 3.70×10−7N )i+( 7.60×10−7N )j=

q( -1.28x i×k - 1.28y j×k - 1.28z k×k)

−( 3.70×10−7N )i+( 7.60×10−7N )j=

q( 1.28xj - 1.28y i )

−( 3.70×10−7N )i+( 7.60×10−7N )j=

q( -1.28y i + 1.28x j)

So comparing comparing coefficients

let compare x axis component

-( 3.70×10−7N )i=-1.28qy i

−3.70×10−7N = -1.28qy

y= -3.7×10^-7/-1.28q

y= -3.7×10^-7/-1.28×-5.90×10^-9)

y=-48.99m/s

y≈-49m/s

Let compare y-axisaxis

7.6×10−7N j = 1.28qx j

7.6×10−7N = 1.28qx

x= 7.6×10^-7/1.28q

x= 7.6×10^-7/1.28×-5.90×10^-9

x=-100.64m/s

a. Then, the velocity of the x component is x= -100.64m/s

b. Also, the velocity component of the y axis is y=-49m/s

c. We will compute

V•F

V=-100.64i -49j

F=−( 3.70×10−7 N )i+( 7.60×10−7 N )j

Note

i.j=j.i=0. Also i.i = j.j =1

V•F is

(-100.64i-49j)•−(3.70×10−7N)i+(7.60×10−7 N )j =

3.724×10^-5 - 3.724×10^-5=0

V•F=0

d. Angle between V and F

V•F=|V||F|Cosx

0=|V||F|Cosx

Cosx=0

x= arccos(0)

x=90°

Since the dot product is zero, from vectors , if the dot product of two vectors is zero, then the vectors are perpendicular to each other

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