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Vo = 10cm = 1000cm^3 = .001m^3

T = 200 C

brass = 19x10^-6 coefficient of linear expansion

V=Vo(1+3 x brass x T)

or

change in V = Vo(3 x brass x T)

= .001m^3(3 x 19x10^-6 x 200 C)

= .0000114 = 1.14x10^-5

% change = change in V/Vo

= .0000114/.001

= .0114 = 1.1%

answer in the book is 1.1x10^-3 %

What I am doing wrong here?

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A radial tire is inflated to a gauge pressure of 35 lb/in^2 at 60 deg. F.

If the temperature increases to 100 deg F while the volume of the

tire remains constant, what is the tire's new pressure?

T1=60 F = 289 K

T2=100 F = 311 K

V1=V2

P1= 35 lb/in^2

Find P2

P1/T1=P2/T2

P2=P1T2/T1

P2 = (35 lb/in^2 x 311 K)/289 K

P2 = 37.7 lb/in^2

Also tried this:

P=pa + pg

14.7 (atm) + 35 lb/in^2

= 49.7 lb/in^2 = absolute pressure

P2 = (49.7 lb/in^2 x 311 K)/289 K

P2 = 53.5 lb/in^2

answer in the book is 39 lb/in^2

any help would be appreciated