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NISA [10]
3 years ago
14

Consider as a system the gas in a vertical cylinder; the cylinder is fitted with a piston on which a number of small weights are

placed. The initial pressure is 200 kPa, and the initial volume of the gas is 0.04 m3 . (a) Let a Bunsen burner be placed under the cylinder, and let the volume of the gas increase to 0.1 m3 while the pressure remains constant. Calculate the work done by the system during this process. (b) Consider the same system and initial conditions, but at the same time that the Bunsen burner is under the cylinder and the piston is rising, remove weights from the piston at a rate such that, during the process, the temperature of the gas remains co
Physics
1 answer:
Amiraneli [1.4K]3 years ago
5 0

Answer:

Explanation:

a ) At constant pressure , work done = P x Δ V

= 200 x 10³ x ( .1 - .04 )

= 12 x 10³ J .

b )

At constant temperature work done

= n RT ln v₂ / v₁

= PV ln v₂ / v₁

= 200 x 10³ x .04 ln .1 / .04

8 x 10³ x .916

= 7.33 x 10³ J .

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125 joules

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Find the total change in the internal energy of a gas that is subjected to the following two-step process. In the first step the
agasfer [191]

Answer:

U = 9253.6 J

Explanation:

It is given that,

Heat transferred in isochoric process, Q_1=5560\ J

Pressure in this process, P_1=3.32\times 10^5\ Pa

In the second step,

It is subjected to isobaric compression until its volume decreases by, V=-7.3\times 10^{-3}\ m^3 (it decreases)

Heat transferred out of the gas, Q_2=1270\ J

In isochoric process, work done by the gas is zero as in this type volume is constant, W_1=0

According to first law of thermodynamics,

\Delta U=Q-W

\Delta U_1=5560\ J.............(1)

In isochoric compression, W_2=P\Delta V

\Delta U_2=1270-3.32\times 10^5\times (-7.3\times 10^{-3})

\Delta U_2=3693.6\ J

Let U is the total change in the internal energy of this gas. So,

U=U_1+U_2

U=5560+3693.6

U = 9253.6 J

So, the total change in the internal energy of this gas is $$9253.6 J. Hence, this is the required solution.

7 0
3 years ago
A particle moving at 10 m/s along the x-axis collides elastically with another particle moving at 5.0 m/s in the same direction
drek231 [11]

Explanation:

It is given that,

Velocity of particle 1, u₁ = 10 m/s

Velocity of particle 2, u₂ = 5 m/s

Let v₁ and v₂ are the final speed of both particles after the collision. Applying the conservation of momentum as :

mu_1+mu_2=mv_1+mv_2

15=v_1+v_2......................(1)

For an elastic collision, the coefficient of restitution is equal to 1 as :

\dfrac{v_2-v_1}{u_1-u_2}=1

\dfrac{v_2-v_1}{5}=1

{v_2-v_1}=5................(2)

On solving equation (1) and (2), we get,

v_1=5\ m/s

v_2=10\ m/s

So, the speeds of particle 1 and particle 2 after the collision is 5 m/s and 10 m/s respectively. Hence, this is the required solution.

8 0
3 years ago
The change in resistance of a metallic conductor at temperature below 0°C is
Aleks [24]

Answer:

A: Linear

Explanation:

Formula for Resistance of a metallic conductor is given as;

R = R_o(1 + αΔT)

Where;

R_o is the original resistance

R is the final resistance after change in temperature

ΔT is change in temperature

α is coefficient of linear expansion

Now, from the formula given, we can see that the change in resistance is directly proportional to the change in temperature.

Thus, the higher the final temperature, the more the change in resistance and the lower the final temperature, the lesser the change in resistance.

Thus, for temperature less than zero, since the change is resistance is directly proportional to the temperature, it means it follows a linear relationship.

5 0
3 years ago
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