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mezya [45]
2 years ago
5

An ideal gas in an isentropic process has an initial pressure of 200 kPa and relative pressure is 3.482; the final pressure is n

ow 320 kPa, find the relative pressure at this state. round to the nearest 3 decimal places
Physics
1 answer:
Nostrana [21]2 years ago
3 0

The relative pressure at this state is determined as 0.455.

<h3>Initial pressure of the ideal gas </h3>

The pressure of the ideal gas given rise to the relative pressure is calculated as follows;

R.P = P2/P1

P2 = P1(R.P)

P2 = 200 kPa x (3.482)

P2 = 696.4 kPa

<h3>New relative pressure</h3>

R.P = (P3)/(P2)

R.P = (320)/(696.4)

R.P = 0.455

Thus, the relative pressure at this state is determined as 0.455.

Learn more about relative pressure here: brainly.com/question/15584931

#SPJ1

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