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True [87]
3 years ago
14

A frictionless piston–cylinder device contains 7 kg of nitrogen at 100 kPa and 250 K. Nitrogen is now compressed slowly accordin

g to the relation PV1.4 = constant until it reaches a final temperature of 450 K. Calculate the work input during this process. The gas constant for nitrogen is R = 0.2968 kJ/kg·K.
Physics
1 answer:
lora16 [44]3 years ago
5 0

Answer: -1038.8 kJ

Explanation:

From the question, we can see that PV^n = constant. And as such, we can deduce that it is a polytropic process. Thus, we can use the polytropic work equation to calculate the needed work input.

from the question we were given

Mass of nitrogen, m = 7kg

initial temperature, T1 = 250k

Final temperature, T2 = 450k

Polytropic index, n = 1.4

Specific gas constant, R = 0.2968kJ/kgK

W = [p2 * v2 - p1 * v1] / 1 - n

W = [m * R * T2 - T1] / 1 - n

W = 7*0.2968*(450 - 250)] / 1 - 1.4

W = [7*0.2968*200] / -0.4

W = 415.52 / -0.4

W = -1038.8 kJ

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Average acceleration on second part of the chunk is given as

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Explanation:

By momentum conservation along x direction we will have

mv_i = \frac{m}{2}v_1 + \frac{m}{2}v_2

so we have

v_1 + v_2 = 2v

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also by energy conservation

\frac{1}{2}(\frac{m}{2})v_1^2 + \frac{1}{2}(\frac{m}{2})v_2^2 - \frac{1}{2}mv^2 = 17 J

\frac{1}{4}m(v_1^2 + v_2^2) - \frac{1}{2}mv^2 = 17

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v_2 = 0.24 m/s

Average acceleration on first part of the chunk is given as

a_1 = \frac{4.44 - 2.34}{0.16}

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Average acceleration on second part of the chunk is given as

a_2 = \frac{0.24 - 2.34}{0.16}

a_2 = -13.125 m/s^2

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