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soldi70 [24.7K]
2 years ago
6

PLEASE HELP ASAP!! CORRECT ANSWER ONLY PLEASE!!

Physics
2 answers:
Maksim231197 [3]2 years ago
8 0

Again, you are correct. It is direction. Speed and velocity have to be measured with something like "mph" or "east at 25 miles per hour" and direction would have to say something like " moved 5 meters SE". The correct choice is distance. Hope that helps!


Shalnov [3]2 years ago
8 0
I believe you’re correct, the answer is:
Distance.

(Its wouldn’t be speed because it isn’t over a certain amount of time, not direction because it doesn’t say anything)
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A. The work done during the process is W = 0

B. The value of heat transfer during the process Q = 442.83 \frac{KJ}{kg}

Explanation:

Given Data

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Final pressure P_{2} = 300 k pa

Vessel is rigid so change in volume of the gas is zero. so that initial volume is equal to final volume.

⇒ V_{1} = V_{2} ------------- (1)

Since volume of the gas is constant so pressure of the gas is directly proportional to the temperature of the gas.

⇒ P ∝ T

⇒ \frac{P_{2} }{P_{1}} = \frac{T_{2} }{T_{1}}

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⇒ T_{2} = \frac{300}{100} × 298

⇒ T_{2} = 894 Kelvin

(A). Work done during the process is given by W = P × (V_{2} -V _{1})

From equation (1), V_{1} = V_{2} so work done W = P × 0 = 0

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Therefore the work done during the process is zero.

Heat transfer during the process is given by the formula Q = m C_{v} ( T_{2} -T_{1} )

Where m = mass of the gas = 1 kg

C_{v} = specific heat at constant volume of nitrogen = 0.743 \frac{KJ}{kg k}

Thus the heat transfer Q = 1 × 0.743 × ( 894- 298 )

⇒ Q = 442.83 \frac{KJ}{kg}

Therefore the value of heat transfer during the process Q = 442.83 \frac{KJ}{kg}

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