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Vinvika [58]
4 years ago
7

Two identical cylinders with a movable piston contain 0.7 mol of helium gas at a temperature of 300 K. The temperature of the ga

s in the first cylinder is increased to 564 K at constant volume by doing work W1 and transferring energy Q1 by heat. The temperature of the gas in the second cylinder is increased to 564 K at constant pressure by doing work W2 while transferring energy Q2 by heat.
A. Find ΔEint, 1, Q1, and W1 for the process at constant volume.
B. Find ΔEint, 2, Q2, and W2 for the process at constant pressure.
Physics
1 answer:
zmey [24]4 years ago
5 0

Answer:

Explanation:

A ) At constant volume :

ΔEint = n Cv x ΔT , n is no of moles , Cv is specific heat at constant volume , ΔT is increase in temperature .

For helium Cv = 3/2 R = 1.5 x 8.3 J = 12.45 J

ΔEint = .7 x 12.45 x ( 564 - 300 )

= 2300.76 J .

W₁ = 0 because volume is constant so work done by gas is zero .

Q₁ = ΔEint = 2300.76 J

B )

At constant pressure

Q₂ = n Cp Δ T , Cp is specific heat at constant pressure .

For monoatomic gas ,

Cp = 5/2 R = 2.5 x 8.3 J = 20.75 J

Q₂ = .7 x 20.75 x 264 J

= 3834.6 J

W₂ = work done by gas

= PΔV = nRΔT

= .8  x 8.3 x 264

= 1752.96 J

ΔEint = Q₂ - W₂

= 3834.6 - 1752.96

= 2081.64 J.

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timurjin [86]

Answer:

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Explanation:

m1=0.20kg

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initial velocity of m1=u1=0.50m/s

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total momentum of the system before collision

Pi=m1u1+m2u2

Pi=0.20kg×0.50m/s+0.30kg×0.40m/s

Pi=0.1kgm/s+0.12kgm/s

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4 years ago
Ted lifts a 10N weight at a height of 1.5 m in 1 second. Johnny lifts a 10N weight at a height of 1.5 m in 2 second. Which state
fenix001 [56]

Answer:

he affirmations the correct ones are: A and D

Explanation:

To find out which expressions are correct, let's calculate the work and power done by each person

the expression for work is W = F and

the expression for power is P = W / t

In this case the work is positive because the beast is in the same direction of displacement.

Ted

        W₁ = 10 1.5

        W₁ = 15 J

        P₁ = 15/1

        P₁ = 15 w

Johnny

the job

         W₂ = 10 1.5

         W₂ = 15 J

the potential

          P₂ = 15/2

          P₂ = 7.5 w

therefore we see that the work of both is the same, but Ted developed twice as much power as Johnny

when resisting the affirmations the correct ones are: A and D

3 0
3 years ago
True or False: Our food has potential energy?
DanielleElmas [232]

Hello User,

Answer is yes.

Explanation

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3 years ago
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A huge cannon is assembled on an airless planet (ignore any effects due to the planet's rotation). The planet has a radius of 5.
Ganezh [65]

Answer:

The projectile's speed as it passes the satellite is 1497.8 m/s.

Explanation:

Given that,

Radius of planet r=5.00\times10^{6}\ m

Mass of planet m=3.95\times10^{23}\ kg

Speed = 2000 m/s

Height = 1000 km

We need to calculate the projectile's speed as it passes the satellite

Using conservation of energy

E_{1}=E_{2}

\dfrac{1}{2}mv_{1}^2+\dfrac{GmM}{r_{1}}=\dfrac{1}{2}mv_{2}^2+\dfrac{GmM}{R+h}

\dfrac{v_{1}^2}{2}+\dfrac{GM}{r_{1}}=\dfrac{v_{2}^2}{2}+\dfrac{GM}{R+h}

-\dfrac{v_{2}^2}{2}=-(\dfrac{GM}{R}-\dfrac{GM}{R+h}-\dfrac{v_{1}^2}{2})

v_{2}^2=v_{1}^2+2GM(\dfrac{1}{R+h}-\dfrac{1}{R})

v_{2}=\sqrt{v_{1}^2+2GM(\dfrac{1}{R+h}-\dfrac{1}{R})}

Put the value into the formula

v_{2}=\sqrt{2000^2+2\times6.67\times10^{-11}\times3.95\times10^{23}(\dfrac{1}{5.00\times10^{6}+1000\times10^{3}}-\dfrac{1}{5.00\times10^{6}})}

v_{2}=1497.8\ m/s

Hence, The projectile's speed as it passes the satellite is 1497.8 m/s.

4 0
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kati45 [8]
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4 0
3 years ago
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