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notsponge [240]
3 years ago
12

An insulated piston–cylinder device contains 100 L of air at 400 kPa and 25°C. A paddle wheel within the cylinder is rotated unt

il 15 kJ of work is done on the air while the pressure is held constant. Determine the final temperature of the air. Neglect the energy stored in the paddle wheel. (Page 172).
Physics
1 answer:
solong [7]3 years ago
7 0

Answer:

T'=70.92°C

Explanation:

Given that

V= 100 L=0.1 m³

P=400 KPa

T=25°C

Work done on the air = 15 KJ

W= -15 KJ

If we assume that air is ideal gas

P V = m R T

R=0.287 KJ/kg.K for air

T= 273 + 25 = 298 K

By putting the values

P V = m R T

400 x 0.1 = m x 0.287 x 298

m=0.46 kg

From first law of thermodynamics

Q= ΔU +W

Insulated piston–cylinder  , Q=0

ΔU  = m Cv  ΔT

   ΔU = - W

Cv = 0.71 KJ/kg.k for air

0.46 x 0.71 x (T' -25) = 15  

T'=70.92 °C

So the final temperature of air is T'=70.92 °C

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Rus_ich [418]

Explanation:

(a)  For an isothermal process, work done is represented as follows.

             W = -nRT ln(\frac{V_{2}}{V_{1}})

Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

             = -12280.82 \times ln (0.09)

             = -12280.82 \times -2.41

             = 29596.78 J

or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = \frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}

              = \frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}

              = \frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}

              = 49.41 kJ

Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

(c)   We know that for an isothermal process,

               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})

                    = 11 atm

Hence, the required pressure is 11 atm.

(d)   For adiabatic process,  

          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

or,       P_{2} = P_{1} (\frac{V_{1}}{V_{2}})^{1.4}

                    = 1 atm (\frac{V_{1}}{\frac{V_{1}}{11}})^{1.4}

                    = 28.7 atm

Therefore, required pressure is 28.7 atm.

6 0
3 years ago
If a planet has a radius 20% greater than that of the Earth but has the same mass as the Earth, what is the acceleration due to
Vaselesa [24]

Answer:

g_2=6.8125 m^2/sec

Explanation:

We know that

Acceleration due to gravity g is given by the formula

g= \frac{GM}{r^2}

G= gravitational constant

M= mass of the Earth

r= radius of the earth

g_1 = GM/r_1^2

Let acc. due to gravity after radius is 20% greater be g_2

then

g_2=GM/r_2^2

=> g1/g2 = (r_2/r_1)^2 => g2 = 9.81/1.2^2 = 6.8125

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the answer is cancer is often named according to what body type it affects

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3 years ago
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2.
Alex_Xolod [135]

Answer:

The work done on the sled by friction, W = - 4593.75 J

Explanation:

Given data,

The combined mass of sled and the boy, m = 75 kg

The displacement of the boy, S = 25 m

The coefficient of the friction, u = 0.25

The frictional force acting on the boy,

                  <em>F = u η</em>

Where,

                        η - is the normal force acting on the boy (mg)

Substituting the values,

                   F = 0.25 x 75 x 9.8

                      = 183.75 N

Since the direction of the frictional force is against the direction of motion

                      F = - 183.75 N

The work done on the sled by friction,

                         W = F x S

                             = - 183.75 x 25

                             = - 4593.75 J

Hence, the work done on the sled by friction, W = - 4593.75 J

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A system consists of two particles. Which of the following scenarios would the force on each particle increase the most? If mult
MrMuchimi

Answer:

c

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