If 18.0 mol of an ideal gas is at 10.0 degree C and the storage canister reads a

Question

If 18.0 mol of an ideal gas is at 10.0 degree C and the storage canister reads a "gauge pressure" of 0.350 atm, what is the volume of the container? Another canister with a volume of 0.820 m^3 which is made of Aluminum contains an ideal gas at 10.0 degree C and the "gauge pressure" reads 0.450 atm. If the gas is heated to 410.0 degree C, neglecting the thermal expansion of the canister, what will the "gauge pressure now read? Also sketch a pV diagram for this process. In part (b) above what is the volume increase in cubic centimeters of the Aluminum canister when it is heated from 10.0 degree C to 410.0 degree C? (The coefficient of volume expansion of Aluminum is 75.0 times 10^-6 per degree Celsius). If we do take this volume expansion of the canister into account, what would be the "gauge pressure"?

Explanation / Answer

a) volume of container = (18 * 8.314 * 283.15)/(0.350 * 1.013 * 105)

                                      = 1.195 m3

b) Here, PV/T = constant

=> 0.450 * 0.820/283.15 = P * 0.820/683.15

=> final pressure = 1.085 atm

c) volume increase = 75 * 10-6 * 0.820 * (410 - 10)

                                 = 0.0246 m3

Also, 0.450 * 0.820/283.15 = P * 0.8446/683.15

=>    Final pressure = 1.054 atm

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