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Vladimir [108]
3 years ago
7

A chamber fitted with a piston contains 1.90 mol of an ideal gas. Part A The piston is slowly moved to decrease the chamber volu

me while the gas temperature is held constant at 296 K. How much work is done by the gas if the final chamber volume is one-third the initial volume
Physics
1 answer:
yarga [219]3 years ago
5 0

Answer:

The work done is 5136.88 J.

Explanation:

Given that,

n = 1.90 mol

Temperature = 296 K

If the initial volume is V then the final volume will be V/3.

We need to calculate the work done

Using formula of work done

W=nRT\ ln(\dfrac{V_{f}}{V_{i}})

Put the value into the formula

W=1.90\times8.314\times296\ ln(\dfrac{\dfrac{V}{3}}{V})

W=1.90\times8.314\times296\ ln(\dfrac{1}{3})

W=−5136.88\ J

The Work done on the system.

Hence, The work done is 5136.88 J.

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A car traveling 90km/hr is 100 m behind a truck traveling 50km/hr. How long will it take the car to reach the truck?
AlexFokin [52]

The faster car behind is catching up/closing the gap/gaining on
the slow truck in front at the rate of (90 - 50) = 40 km/hr.

At that rate, it takes (100 m) / (40,000 m/hr) = 1/400 of an hour
to reach the truck.

(1/400 hour) x (3,600 seconds/hour) = 3600/400 = <em>9 seconds</em>, exactly 


4 0
3 years ago
Read 2 more answers
What is the power of a parallel circuit with a resistance of 1,000 and a current of 0.03 A?
Kobotan [32]

i squared r = 0.03x0.03x1000=3x0.03x10=.9W

7 0
3 years ago
A 1.11 kg piece of aluminum at 78.3 c is put into a glass with 0.210 kg of water at 15.0
ValentinkaMS [17]

M = mass of aluminium = 1.11 kg

c_{a} = specific heat of aluminium = 900

T_{ai} = initial temperature of aluminium = 78.3 c

m = mass of water = 0.210 kg

c_{w} = specific heat of water = 4186

T_{wi} = initial temperature of water = 15 c

T = final equilibrium temperature = ?

using conservation of heat

Heat lost by aluminium = heat gained by water

M c_{a} (T_{ai} - T) = m c_{w} (T - T_{wi} )

(1.11) (900) (78.3 - T) = (0.210) (4186) (T - 15)

T = 48.7 c

7 0
3 years ago
Air enters a nozzle steadily at 2.21 kg/m3 and 20 m/s and leaves at 0.762 kg/m3 and 150 m/s. If the inlet area of the nozzle is
saveliy_v [14]

Answer:

a) The mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle is 23.6 cm².

Explanation:

a) The mass flow rate through the nozzle can be calculated with the following equation:

\dot{m_{i}} = \rho_{i} v_{i}A_{i}

Where:

v_{i}: is the initial velocity = 20 m/s

A_{i}: is the inlet area of the nozzle = 60 cm²  

\rho_{i}: is the density of entrance = 2.21 kg/m³

\dot{m} = \rho_{i} v_{i}A_{i} = 2.21 \frac{kg}{m^{3}}*20 \frac{m}{s}*60 cm^{2}*\frac{1 m^{2}}{(100 cm)^{2}} = 0.27 kg/s  

Hence, the mass flow rate through the nozzle is 0.27 kg/s.

b) The exit area of the nozzle can be found with the Continuity equation:

\rho_{i} v_{i}A_{i} = \rho_{f} v_{f}A_{f}

0.27 kg/s = 0.762 kg/m^{3}*150 m/s*A_{f}

A_{f} = \frac{0.27 kg/s}{0.762 kg/m^{3}*150 m/s} = 0.00236 m^{2}*\frac{(100 cm)^{2}}{1 m^{2}} = 23.6 cm^{2}

Therefore, the exit area of the nozzle is 23.6 cm².

I hope it helps you!                                                                  

5 0
3 years ago
Read 2 more answers
A strontium vapor laser beam is reflected from the surface of a CD onto a wall. The brightest spot is the reflected beam at an a
sladkih [1.3K]

Answer:

d=1.29*10^{-6}m

Explanation:

From the question we are told that:

Distance of wall from CD D=1.4

Second bright fringe y_2= 0.803 m

Let

Strontium vapor laser has a wavelength \lambda= 431 nm=>431 *10^{-9}m

Generally the equation for Interference is mathematically given by

y=frac{n*\lambda*D}{d}

Where

d=\frac{n*\lambda*D}{y}

d=\frac{2*431 *10^{-9}m*1.4}{0.803}

d=1.29*10^{-6}m

8 0
2 years ago
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