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just olya [345]
3 years ago
15

Quantity A of an ideal gas is at absolute temperature T, and a second quantity B of the same gas is at absolute temperature 2T.

Heat is added to each gas, and both gases are allowed to expand isothermally.If both gases were initially at the same absolute temperature, would they still undergo the same entropy change?No, gas A would undergo the greater entropy change.No, gas B would undergo the greater entropy change.Yes, both gases would have the same entropy.
Physics
1 answer:
Ira Lisetskai [31]3 years ago
5 0

Answer: Yes both gases would have the same entropy.

Explanation:

The formula for the change in the entropy is as follows,

\Delta S = \frac{Q}{T}

Here, \Delta S is the change in the entropy, Q is the heat transfer and T is the temperature.

If the temperature of the system increases then there will be increase in the entropy as the randomness of the system increases.

In the given problem, if the both gases were initially at the same absolute temperature. Then there will be same entropy change in both gases.

Therefore, yes both gases would have the same entropy.

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Answer:

The ball would hit the floor approximately 0.55\; \rm s after leaving the table.

The ball would travel approximately 5.5\; \rm m horizontally after leaving the table.

(Assumption: g = 9.81\; \rm m \cdot s^{-2}.)

Explanation:

Let \Delta h denote the change to the height of the ball. Let t denote the time (in seconds) it took for the ball to hit the floor after leaving the table. Let v_0(\text{vertical}) denote the initial vertical velocity of this ball.

If the air resistance on this ball is indeed negligible:\displaystyle \Delta h = -\frac{1}{2}\, g\, t^{2} + v_0(\text{vertical}) \cdot t.

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