A mixture of gases A2 and B2 are introduced to a slender metal cylinder that has one end closed and the other fitted with a piston that makes a gas-tight seal so that the gases are a closed system. The cylinder is submerged in a large beaker of water whose temperature is 25∘C, and a spark is used to trigger a reaction in the cylinder. At the completion of the reaction, the piston has moved downward, and the temperature of the water bath has increased to 28∘C. If we define the system as the gases inside the cylinder, which of the following best describes the signs of q, w, and ΔE for this reaction?

(a) q < 0, w < 0, ΔE < 0
(b) q < 0, w > 0, ΔE < 0
(c) q < 0, w > 0, the sign of ΔE cannot be determined from the information given
(d) q > 0, w > 0, ΔE > 0
(e) q > 0, w < 0, the sign of ΔE cannot be determined from the information given

After the completion of the reaction, temperature of the water bath is increased that means heat is released during the reaction and flows out of the system.

If heat is absorbed by the system, then q is indicated by positive sign and if heat is released by the system, then q is indicated by negative sign.

As in the given case, heat is released by the system, so sign of q is negative, or q < 0

Determination of sign of w

After the completion of the reaction, piston moved downward, that means volume of the system decreases or compression occur. During the compression, work is done on the system.

if work is done on the system, sign of w is positive.

If work is done by the system, sign of w is negative.

In the given case, work is done on the system, therefore sign of w is positive, or w > 0

Determination of sign of ΔE

Relationship between ΔE, q and w is given by first law of thermodynamics:

ΔE = q + w

In this case, q is positive and w is negative, so the sign of ΔE depends of magnitude of q and w. As magnitude of w and q cannot be determined in this case, thus, the given information is insufficient for the determination of sign of ΔE.

So, among the given option, option c is correct.

q < 0, w > 0, the sign of ΔE cannot be determined from the information given

ajeigbeibraheem ajeigbeibraheem · Answer 2 · 2021-11-20T23:17:22+01:00

The best option that describes the signs of q, w, and ΔE for this reaction is q < 0, w > 0, the sign of ΔE cannot be determined from the information given.

Option C is correct.

From the given information, if we examine the mixture of the gases A2 and B2 and take them to be the system, and the slender metal cylinder as well as the water bath to be the surroundings.

Then, we can infer that:

if there is a movement of heat flow(q) into the system, it signifies a positive sign
if there is an outflow of heat (q) out of the system, it signifies a negative sign

Hence, when the reaction is said to be completed, we are being informed from the question that the temperature increased. This implies that there is an outflow and release of heat away from the system. As such, when heat flows away from the system, the sign (q) will be negative.

For the work done (w):

suppose work is done on a system show that the sign will be positive;
and suppose the work is done by the system, then it will be negative.

When the reaction goes into completion, the movement of the piston downward indicates that there is a decrease in the volume of the system.

As such, work is done on the system, hence, the work done (w) is positive.

For the change in the internal energy ΔE. If we look at the first law of thermodynamics, we know that:

ΔE = q + w

here,

we know q to be -ve and w to be +ve;

However, the sign of ΔE largely depends on the magnitude of q and w. We can infer that there is little information given for the determination of ΔE

Therefore, we can conclude that q < 0 since it is negative, w > 0 since it is positive and the sign of ΔE cannot be determined from the information given.

Learn more about the first law of thermodynamics here:

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