Two punds of water vapor at 30 psia fill the 4ft3 left chmaber of a partitioned system. the right chmaber has twice the volume of the left and is initially evacuated. detrmine the pressure of water after the paertiion has been removed nd enough heat has been transfered so that the temperature of the water is 40f.

pressure of water will be 49.7 atm

Explanation:

given data

pressure = 30 psi = 2.04 atm

water = 2 pound = 907.18

mole of water vapor = 907.19 /2 = 50.4 mole

volume = 4 ft³ = 113.2 L

temperature = 40 F = 277.59 K

to find out

pressure of water

solution

we will apply here ideal gas condition

that is

PV = nRT  1

put here all value and here R = 0.0821 , T temperature and V volume and P pressure and n is no of mole

and we get here temperature

PV = nRT  

2.04 × 113.2 = 50.4×0.0821×T

solve it and we get

T = 55.8 K

so we have given right chamber has twice the volume of the left chamber i.e

volume = twice of volume + volume

volume = 2(113.2) + 113.2

volume = 339.6 L

so from equation 1 pressure will be

PV = nRT

P(339.6) = 50.4 × ( 0.0821) × (277.59)

P = 3.3822 atm = 49.7 atm

so pressure of water will be 49.7 atm

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total distance/total time

(85 + 82)/(5+3)

167/8

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