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Alexandra [31]
3 years ago
10

A sample of 4.50 g of methane occupies 12.7 dm3 at 310 K. (a) Calculate the work done when the gas expands isothermally against

a constant external pressure of 200 Torr until its volume has increased by 3.3 dm3. (b) Calculate the work that would be done if the same expansion occurred reversibly.
Physics
1 answer:
valkas [14]3 years ago
7 0

Answer:

(A) Work done will be 87.992 KJ

(B) Work done will be 167.4 KJ            

Explanation:

We have given mass of methane m = 4.5 gram = 0.0045 kg

Volume occupies V_1=12.7dm^3=12.7liters

And volume is increased by 3.3dm^3 so V_2=12.7+3.3=16liters

Temperature T = 310 K

Pressure is given as 200 Torr = 26664.5 Pa

(a) At constant pressure work done is given by

W=P(V_2-V_1)=26664.5\times (16-12.7)=87992.85J=87.992kj

(b) At reversible process work done is given by W=nRTln\frac{V_2}{V_1}

We have given mass = 4.5 gram

Molar mass of methane = 16

So number of moles n=\frac{mass\ in\ gram}{mol;ar\ mass}=\frac{4.5}{16}=0.28125

So work done W=0.28125\times 8.314\times 310ln\frac{16}{12.7}=167.4J

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If a seismic wave has a period of 0.0202s, find the frequency of the wave.
maw [93]

Answer:

49.5 Hz.

Explanation:

From the question given above, the following data were obtained:

Period (T) = 0.0202 s

Frequency (f) =?

The frequency and period of a wave are related according to the following equation:

Frequency (f) = 1 / period (T)

f = 1/T

With the above formula, we can obtain the frequency of the wave as follow:

Period (T) = 0.0202 s

Frequency (f) =?

f = 1/T

f = 1/0.0202

f = 49.5 Hz

Therefore the frequency of the wave is 49.5 Hz.

7 0
3 years ago
A very strong, but inept, shot putter puts the shot straight up vertically with an initial velocity of 11.0 m/s. how long does h
noname [10]
The shot putter should get out of the way before the ball returns to the launch position.

Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.

The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 =  0.45 s

t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.

Answer: 0.45 s
7 0
3 years ago
#2 A car accelerates from rest at 4 m/s^2. What is the velocity of the car after 4 second?
solmaris [256]
You use the equation Velocity = Acceleration X Time. 4x4=16m/s.


The car travels 18m in 3 seconds.
3 0
3 years ago
A motor has an internal resistance of 12.1 Ω. The motor is in a circuit with a current of
kotykmax [81]

Answer:

Explanation:

V = I * R

V = 4 * 12.1 = 48.4 v

7 0
2 years ago
4. A rock is thrown from the edge of the top of a 100 m tall building at some unknown angle above the horizontal. The rock strik
nikklg [1K]

Answer:

Explanation:

Let the velocity of projectile be v and angle of throw be θ.

The projectile takes 5 s to touch the ground during which period it falls vertically by 100 m

considering its vertical displacement

h = - ut +1/2 g t²

100 = - vsinθ x 5 + .5 x 9.8 x 5²

5vsinθ =  222.5

vsinθ = 44.5

It covers 160 horizontally in 5 s

vcosθ x 5 = 160

v cosθ = 32

squaring and adding

v²sin²θ +v² cos²θ = 44.4² + 32²

v² = 1971.36 + 1024

v = 54.73 m /s

5 0
3 years ago
Read 2 more answers
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