Question

Pressure of an ideal gas decreases to 5% at 5°C
of its original value of 5 atm at 283°C. What is the
percentage change in its density?​

Answer 1

Answer:

Hey Dear,

◆ Answer -

∆d/d % = -5 %

◆ Explanation -

# Given -

P = 5 atm

∆P = - 5% P

# Solution -

Final pressure is given by -

∆P = P' - P

-5/100 P = P' - P

P' = 95/100 P

P/P' = 100/95

According to Boyle's law,

PV = P'V'

V'/V = P/P'

V'/V = 100/95

Initial and final density of gas is related as -

d/d' = (M/V) / (M/V')

d/d' = V'/V

d/d' = 100/95

d' = 95/100 d

Therefore, percentage change in density is -

∆d/d % = (d'-d) / d × 100

∆d/d % = (95/100d - d) / d × 100

∆d/d % = -5 %

Therefore, density of the gas decreases by 5 % during the process.

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Explanation:

Answer 2

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