urse Home https:// essonmastenn cblenD-89583508 Physics 222 Homework 8 (Chapter

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urse Home https:// essonmastenn cblenD-89583508 Physics 222 Homework 8 (Chapter 18) Problem 18.58 c) 16of18 Part A The 50 kg circular lead piston shown in (Figure 1) floats on 0 33 mol of compressed air What is te piston hoight it the temperature is Express your answer with the appropriate units 300 hi 065 m Previous Aaswers Figure 1ol 1 Correct 1 atm Part B Now far does the piston meove & the temperahure is increased by 100 C Express your answer with the appropriate unts 50 kg Pant 30 .h-14.286 cm 10 cm Submit e here to search

Explanation / Answer

Pressure due to the weight of piston
= force /Area = [(50*9.81) /{(Pi/4)*(0.1)2}] = 62.452 kPa
And atmospheric pressure = 1 atm = 101.324 kPa
Total pressure = 62.452 + 101.324 = 163.772 kPa
(B) Since the temperature is raised and piston is free to move therefore the pressure will remain same.
hence
PV = nRT
V = nRT /P = 0.33*8.314*(373) / 163.772*103
V = 6.249*10-3 m3
Volume , V = (Pi/4)d2h2
(Pi/4)d2h2 =  6.249*10-3
h2 = 0.796 m
h2 - h1 = 0.796 - 0.65 = 0.146 m

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