How much energy is wasted if a 100 W device is left on for 20 hours

A sample of monatomic ideal gas occupies 5.00 L at atmospheric pressure and 300 K (point A). It is warmed at constant volume to

leonid [27]

Answer:

(a) 0.203 moles

(b) 900 K

(c) 900 K

(d) 15 L

(e) A → B, W = 0, Q = Eint = 1,518.91596 J

B → C, W = Q ≈ 1668.69974 J Eint = 0 J

C → A, Q = -2,531.5266 J, W = -1,013.25 J, Eint = -1,518.91596 J

(g) ∑Q = 656.089 J, ∑W = 655.449 J, ∑Eint = 0 J

Explanation:

At point A

The volume of the gas, V₁ = 5.00 L

The pressure of the gas, P₁ = 1 atm

The temperature of the gas, T₁ = 300 K

At point B

The volume of the gas, V₂ = V₁ = 5.00 L

The pressure of the gas, P₂ = 3.00 atm

The temperature of the gas, T₂ = Not given

At point C

The volume of the gas, V₃ = Not given

The pressure of the gas, P₃ = 1 atm

The temperature of the gas, T₂ = T₃ = 300 K

(a) The ideal gas equation is given as follows;

P·V = n·R·T

Where;

P = The pressure of the gas

V = The volume of the gas

n = The number of moles present

R = The universal gas constant = 0.08205 L·atm·mol⁻¹·K⁻¹

n = PV/(R·T)

∴ The number of moles, n = 1 × 5/(0.08205 × 300) ≈ 0.203 moles

The number of moles in the sample, n ≈ 0.203 moles

(b) The process from points A to B is a constant volume process, therefore, we have, by Gay-Lussac's law;

P₁/T₁ = P₂/T₂

∴ T₂ = P₂·T₁/P₁

From which we get;

T₂ = 3.0 atm. × 300 K/(1.00 atm.) = 900 K

The temperature at point B, T₂ = 900 K

(c) The process from points B to C is a constant temperature process, therefore, T₃ = T₂ = 900 K

(d) For a constant temperature process, according to Boyle's law, we have;

P₂·V₂ = P₃·V₃

V₃ = P₂·V₂/P₃

∴ V₃ = 3.00 atm. × 5.00 L/(1.00 atm.) = 15 L

The volume at point C, V₃ = 15 L

(e) The process A → B, which is a constant volume process, can be carried out in a vessel with a fixed volume

The process B → C, which is a constant temperature process, can be carried out in an insulated adjustable vessel

The process C → A, which is a constant pressure process, can be carried out in an adjustable vessel with a fixed amount of force applied to the piston

(f) For A → B, W = 0,

Q = Eint = n·cv·(T₂ - T₁)

Cv for monoatomic gas = 3/2·R

∴ Q = 0.203 moles × 3/2×0.08205 L·atm·mol⁻¹·K⁻¹×(900 K - 300 K) = 1,518.91596 J

Q = Eint = 1,518.91596 J

For B → C, we have a constant temperature process

Q = n·R·T₂·㏑(V₃/V₂)

∴ Q = 0.203 moles × 0.08205 L·atm/(mol·K) × 900 K × ln(15 L/5.00 L) ≈ 1668.69974 J

Eint = 0

Q = W ≈ 1668.69974 J

For C → A, we have a constant pressure process

Q = n·Cp·(T₁ - T₃)

∴ Q = 0.203 moles × (5/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -2,531.5266 J

Q = -2,531.5266 J

W = P·(V₂ - V₁)

∴ W = 1.00 atm × (5.00 L - 15.00 L) = -1,013.25 J

W = -1,013.25 J

Eint = n·Cv·(T₁ - T₃)

Eint = 0.203 moles × (3/2) × 0.08205 L·atm/(mol·K) × (300 K - 900 K) = -1,518.91596 J

Eint = -1,518.91596 J

(g) ∑Q = 1,518.91596 J + 1668.69974 J - 2,531.5266 J = 656.089 J

∑W = 0 + 1668.69974 J -1,013.25 J = 655.449 J

∑Eint = 1,518.91596 J + 0 -1,518.91596 J = 0 J

5 0

3 years ago

The copper wire to the motor is 6.0 mm in diameter and 1.1 m long. How far doesan individual electron travel along the wire whil

lesya [120]

Answer:

0.306mm

Explanation:

The radius of the conductor is 3mm, or 0.003m

The area of the conductor is:

A = π*r^2 = π*(.003)^2 = 2.8*10^-5 m^2

The current density is:

J = 130/2.8*10^-5 = 4.64*10^6 A/m

According to the listed reference:

Vd = J/(n*e) = 4.64*10^6 / ( 8.46*10^28 * 1.6*10^-19 ) = 0.34*10^-6 m/s = 0.34mm/s

The distance traveled is:

x = v*t = 0.34 * .90 = 0.306 mm

5 0

4 years ago

the net force on a vehicle is accelerating at a rate of 1.5 milliseconds is 1800 Newtons. What is the mass of the vehicles neare

Goryan [66]

Answer:

The mass is 1200 kilograms

Explanation:

Because Force is equal to mass times acceleration (F=m×a)

F=m×a

1800N=?×1.5

1800÷1.5=1200

1800N=1200Kg×1.5

7 0

4 years ago

Ali is whirling a 2.0 kg bunch of bananas in a circular path having a radius of 0.50 m. The bananas complete 2 revolutions every

skad [1K]

Answer:

1) 2.467 N

2) a) 0.248m

b) 2.3π rad/sec

Explanation:

Given data:

mass of Banana bunch ( m ) = 2.0 kg

radius of circular path ( R ) = 0.5 m

number of revolutions completed = 2

Time to complete 2 revolutions = 6 seconds

1) Determine the force to keep the motion constant for one complete revolution in every 4 seconds

F = mv^2 / r ----- ( 1 )

where V = 2πR/T

where : R = 0.5 , m = 2, T = 4 seconds

Insert values into equation 1

F = 2 * 4π^2 * 0.5/4^2

= 2.467 N

2a) Calculate the maximum distance of coin from center

angular velocity ( w ) = v/r

coefficient of static friction ( μ ) = 0.25

$F_{c} = u mg$ ---- ( 1 )

mv^2/r = μmg --- ( 2 ) cancelling the mass on both sides eqn 2 becomes

v^2 = μ*g*r

dividing both sides of equation by r^2

w^2 = μ*g/r

hence determine distance ( r ) of coin from center

r = 0.25 * 9.81 / π^2 = 0.248 m

2b ) determine the maximum speed of rotation of the turntable for the coin to move relative to the turntable without slipping

distance coin is placed ( r ) = 4.7 cm = 0.047 m

find speed of rotation ( w )

w^2 = μ*g/r

w = √ 0.25 * 9.81/ 0.047

= 7.2236 rad/secs ≈ 2.3π rad/sec

3 0

3 years ago

A 62.0 kg water skier at rest jumps from the dock into a 775 kg boat at rest on the east side of the dock. If the velocity of th

Dominik [7]

Answer:

v = 0.33 m/s

Explanation:

The final velocity of the skier and boat can be calculated by conservation of the lineal momentum:

$m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f} + m_{2}v_{2f}$

<u>Where:</u>

m1: is the mass of the water skier = 62.0 kg

m2: is the mass of the boat = 775 kg

v1i: is the initial velocity of the water skier = 4.50 m/s

v1f: is the final velocity of the water skier = ?

v2i: is the initial velocity of the boat = 0

v2f: is the final velocity of the boat = ?

Since the final velocity of the skier is the same that the final velocity of the boat, we have:

$m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{f} + m_{2}v_{f}$

$m_{1}v_{1i} + 0 = v_{f}(m_{1} + m_{2})$

$v_{f} = \frac{62.0 kg*4.50 m/s}{(62.0 kg + 775 kg)}$

$v_{f} = 0.33 m/s$

Therefore, the final velocity of the skier and boat is 0.33 m/s.

I hope it helps you!

5 0

4 years ago