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Stells [14]
4 years ago
13

Suppose 1.4 mol of an ideal gas is taken from a volume of 2.5 m3 to a volume of 1.0 m3 via an isothermal compression at 27°C. (a

) How much energy is transferred as heat during the compression, and (b) is the transfer to or from the gas?
Physics
1 answer:
zysi [14]4 years ago
6 0

Answer:

Part a)

Q = 3198 J

Part b)

It is compression of gas so this is energy transferred to the gas

Explanation:

Part a)

Energy transfer during compression of gas is same as the work done on the gas

In isothermal process work done is given by the equation

W = nRT ln(\frac{V_2}{V_1})

now we know that

n = 1.4 moles

T = 27 degree C = 300 K

V_2 = 2.5 m^3

V_1 = 1 m^3

now we have

W = (1.4)(8.31)(300)(ln\frac{2.5}{1})

Q = 3198 J

Part b)

It is compression of gas so this is energy transferred to the gas

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hope this will help you more

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With manual transmission, use your right foot for the brake and accelerator and left foot for the clutch true or false
GaryK [48]

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4 0
4 years ago
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A supply plane needs to drop a package of food to scientists working on a glacier in Greenland. The plane flies 110 m above the
zysi [14]

Answer:

The distance is 709.5 m.

Explanation:

Given that,

Speed = 150 m/s

Distance = 110 m

Suppose, How far short of the target should it drop the package?

We need to calculate the time

Using equation of motion

s=ut+\dfrac{1}{2}gt^2

t^2=\dfrac{2s}{g}

Where, g = acceleration due to gravity

t = time

Put the value into the formula

t=\sqrt{\dfrac{2\times110}{9.8}}

t=4.73\ sec

We need to calculate the distance

Using formula of distance

d= vt

Put the value into the formula

d=150\times4.73

d=709.5\ m

Hence, The distance is 709.5 m.

8 0
3 years ago
A projectile is shot from the edge of a cliff above the ground level with initial velocity of at an angle with the horizontal. (
Xelga [282]

Answer:

t = √2y/g

Explanation:

This is a projectile launch exercise

a) The vertical velocity in the initial instants (v_{oy} = 0) zero, so let's use the equation

     y =v_{oy} t -1/2 g t²

     y= - ½ g t²

     t = √2y/g

b) Let's use this time and the horizontal displacement equation, because the constant horizontal velocity

     x = vox t

     x = v₀ₓ √2y/g

c) Speeds before touching the ground

     vₓ = vox = constant

     v_{y} = v_{oy} - gt

     v_{y} = 0 - g √2y/g

    v_{y}  = - √2gy

    tan θ = Vy / vx

    θ = tan⁻¹ (vy / vx)

    θ = tan⁻¹ (√2gy / vox)

d) The projectile is higher than the cliff because it is a horizontal launch

6 0
4 years ago
Two loudspeakers S1 and S2, 2.20 m apart, emit the same single-frequency tone in phase at the speakers. A listener L is located
seropon [69]

Answer:

the lowest possible frequency of the emitted tone is 404.79 Hz

Explanation:

   Given the data in the question;

S₁ ←  5.50 m → L

↑

2.20 m

↓

S₂

We know that, the condition for destructive interference is;

Δr = ( 2m + \frac{1}{2} ) × λ

where m = 0, 1, 2, 3 .......

Path difference between the two sound waves from the two speakers is;

Δr = √( 5.50² + 2.20² ) - 5.50

Δr = 5.92368 - 5.50

Δr = 0.42368 m

v = f × λ

f = ( 2m + \frac{1}{2})v / Δr

m = 0, 1, 2, 3, ....

Now, for the lowest possible frequency, let m be 0

so

f = ( 0 + \frac{1}{2})v / Δr

f = \frac{1}{2}(v) / Δr

we know that speed of sound in air v = 343 m/s

so we substitute

f = \frac{1}{2}(343) / 0.42368

f = 171.5 / 0.42368

f = 404.79 Hz

Therefore, the lowest possible frequency of the emitted tone is 404.79 Hz

5 0
3 years ago
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