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belka [17]
3 years ago
12

0.0884 moles of a diatomic gas

Physics
2 answers:
Sloan [31]3 years ago
7 0

Answer:

W = - 118.24 J (negative sign shows that work is done on piston)

Explanation:

First, we find the change in internal energy of the diatomic gas by using the following formula:

\Delta\ U = nC_{v}\Delta\ T

where,

ΔU = Change in internal energy of gas = ?

n = no. of moles of gas = 0.0884 mole

Cv = Molar Specific Heat at constant volume = 5R/2 (for diatomic gases)

Cv = 5(8.314 J/mol.K)/2 = 20.785 J/mol.K

ΔT = Rise in Temperature = 18.8 K

Therefore,

\Delta\ U = (0.0884\ moles)(20.785\ J/mol.K)(18.8\ K)\\\Delta\ U = 34.54\ J

Now, we can apply First Law of Thermodynamics as follows:

\Delta\ Q = \Delta\ U + W

where,

ΔQ = Heat flow = - 83.7 J (negative sign due to outflow)

W = Work done = ?

Therefore,

-83.7\ J = 34.54\ J + W\\W = -83.7\ J - 34.54\ J\\

<u>W = - 118.24 J (negative sign shows that work is done on piston)</u>

Mandarinka [93]3 years ago
7 0

Answer:

-49.2

Explanation:

Trust me bro

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Answer:

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Explanation:

Given:-

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- The angle of inclination, θ = 20°

- The coefficient of friction of inclined surface, u = 0.3

Find:-

a) The magnitude of tension in the cable

b) The acceleration of the system

Solution:-

- We will first draw a free body diagram for both the blocks. The vertically hanging block of mass m = 10 kg tends to move "upward" when the system is released.

- The block experiences a tension force ( T ) in the upward direction due the attached cable. The tension in the cable is combated with the weight of the vertically hanging block.

- We will employ the use of Newton's second law of motion to express the dynamics of the vertically hanging block as follows:

                        T - m*g = m*a\\\\  ... Eq 1

Where,

              a: The acceleration of the system

- Similarly, we will construct a free body diagram for the inclined block of mass M = 100 kg. The Tension ( T ) pulls onto the block; however, the weight of the block is greater and tends down the slope.

- As the block moves down the slope it experiences frictional force ( F ) that acts up the slope due to the contact force ( N ) between the block and the plane.

- We will employ the static equilibrium of the inclined block in the normal direction and we have:

                        N - M*g*cos ( Q )= 0\\\\N = M*g*cos ( Q )

- The frictional force ( F ) is proportional to the contact force ( N ) as follows:

                        F = u*N\\\\F = u*M*g *cos ( Q )

- Now we will apply the Newton's second law of motion parallel to the plane as follows:

                       M*g*sin(Q) - T - F = M*a\\\\M*g*sin(Q) - T -u*M*g*cos(Q)  = M*a\\ .. Eq2

- Add the two equation, Eq 1 and Eq 2:

                      M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g = a* ( M + m )\\\\a = \frac{M*g*sin ( Q ) - u*M*g*cos ( Q ) - m*g}{M + m} \\\\a = \frac{100*9.81*sin ( 20 ) - 0.3*100*9.81*cos ( 20 ) - 10*9.81}{100 + 10}\\\\a = \frac{-39.12977}{110} = -0.35572 \frac{m}{s^2}

- The inclined block moves up ( the acceleration is in the opposite direction than assumed ).

- Using equation 1, we determine the tension ( T ) in the cable as follows:

                     T = m* ( a + g )\\\\T = 10*( -0.35572 + 9.81 )\\\\T = 94.54 N

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