Review of Information after Exam 3

Concept Checker As a volume of a confined gas decreases at a constant temperature, the pressure exerted by the gas… Decreases Increases Stays the same fluctuates

Answer: Increases P = nRT/V

Practice Problem 1: A sealed container has a volume of.060 m^3 and contains 8 grams of molecular oxygen (O2) which has molecular mass of 32 u. The gas is at a temperature of 345 K. What is the absolute pressure of the nitrogen gas?

Solution: PV=nRT N = 8 /32 =.25 P =.25(8.31)(345) /.060 P = 1.2 * 10 ^ 4 Pa

Practice Problem 2: When a supply of hydrogen gas is held in a 4 liter container at 320 K, it exerts a pressure of 800 torr. The supply is moved to a 2 liter container, and cooled to 160 K. What is the new pressure of the confined gas?

Solution: 800 Torr

Practice 3: Assuming that O2 behaves as an ideal gas and determine the rms speed of the nitrogen molecules when the temperature of the air is 333K.

Solution:.5mv^2 = 3/2 kT M = mass per mol / Na M = 32 / * 10 ^ 23 = 5.31 * 10 ^ -23 V= (3* 1.38 * 10 ^23)(333) / 5.31 * 10 ^ - 23 ^.5 V= 16 m/s

Practice 4: An ideal gas absorbs 238 J of heat as it performs 845 J of work. What is the resulting change in temperature if there are 2.4 moles of an ideal gas in the system.

Solution: Q= +238 W= +845 U= 238 – 845 = J U = 3/2 nRT T= 3/2 U / nR T= -20

Practice 5: A system containing an ideal gas at a constant pressure of 2.2 * 10^5 Pa gains 2431 J of heat. During the process, the internal energy of the system increases by 1243J. What is the change in volume of the gas?

Solution: P = 2.2 * 10 ^ 5 Q= 2431 U = 1243 W= Q – U = 1188 w=pv V= 1188 / 2.2 * 10 ^ 5 = 5.4 * 10 ^ -3

Practice Problem 6: Carnot engine operates between hot and cold reservoirs with temperatures 600 C to -25 C. If the engine performs 2300 J of work per cycle, how much heat is extracted per cycle from the hot reservoir?

Solution: E = 1 - Qc / Qh E= 1 – Tc / Th W = e Qh Sooo  Qh = w/e = w/ ( 1- Tc / Th ) 2300 / ( 1 – ( / )) Qh = 913.2