Chemistry 12th Edition

Published by McGraw-Hill Education
ISBN 10: 0078021510
ISBN 13: 978-0-07802-151-0

Chapter 5 - Gases - Questions & Problems - Page 217: 5.51

Answer

$ 1.5 atm$

Work Step by Step

Given $T_{Ne}=30°C=303K$ $T_{N_{2}}=20°C=293K$ $P_{N_{2}}=1 atm$ Ideal gas constant$ R=0.0821 atm L/Kmol$ Molar mass of $ M_{Ne}=20.18 g/mol$ Molar mass of $M_{N_{2}}= 2\times14.007g/mol = 28.014 g/mol $ Density of nitrogen $d=\frac{m}{V}$ PV=nRT ($n=\frac{m}{M}$) So $d=\frac{PM}{RT}$ For nitrogen $d=\frac{1*28.014}{0.082*293} =1.2 g/L $ Both gases have the same density. So pressure of neon gas is $P_{Ne}=\frac{dRT}{M}$ $P_{Ne}=\frac{1.2*0.082*30}{20.18}= 1.5 atm $
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