The inside of the cylindrical can shown above has cross-sectional area 0.005 m2

Question

The inside of the cylindrical can shown above has cross-sectional area 0.005 m2 and length 0.15 m. The can is filled with an ideal gas and covered with a loose cap. The gas is heated to 363 K and some is allowed to escape from the can so that the remaining gas reaches atmospheric pressure (1.0 times 105 Pa). The cap is now tightened, and the gas is cooled to 298 K. What is the pressure of the cooled gas? Determine the upward force exerted on the cap by the cooled gas inside the can. If the cap develops a leak; how many moles of air would enter the can as it reaches a final equilibrium at 298 K and atmospheric pressure? (Assume that air is an ideal gas.)

Explanation / Answer

a) PV = nRT since V is constant, P1*V = nRT1, P2*V=nRT2 P2/P1= T2/T1 P2= (T2/T1)P1 = (298/363)(10^5) = 82094 Pa b) Upward force = P2*(area of lid) = 82094*.005 = 410 N c) It already has n = PV/RT = (82094)(.005)(.15)/(8.31*298) = .025 mole after equilibrium is reached, it has n' = (10^5)(.005)(.15)/(8.31*298) = .030 mole so .005 mole of air leaks in.

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