100A QUIZ 5
Solution. The change in length between cold and hot conditions is
DL = aL0 DT = [17 x 10-6(Cº)-1](35.0m)[35.0°C - (-20.0°C)] = 3.27 x 10-2m = 3.27cm.
If 4.00m3 of a gas initially at STP is placed under a pressure of 5.00atm, the temperature of gas rises to 40.0ºC. what is the volume?
Solution. For the two states of the gas we can write
P1V1 = nRT1 and P2V2 = nRT2, which can be combined to give
(P2/P1)(V2/V1) = T2/T1;
(5.00atm/1.00atm)(V2/4.00m3) = (313K/273K) , which gives V2 = 0.917m3