Question

1) A cylinder of length L = 1.00 m and inside radius r = 15.0 cm is filled with 5.00 moles of xenon gas. 1 A disc of the same radius and negligible thickness, with a mass of m = 1.75 kg, is slowly lowered into the cylinder, compressing the gas. Then the rest of the cylinder is filled with 3.00 moles of helium gas, and sealed. Both gasses are maintained at a constant temperature of T = 100.0°C. Assume that the disc fits tightly against the sides of the cylinder so that no gas can leak from the bottom to the top, or vice versa, but that the disc is free to slide frictionlessly in the cylinder. Also assume that both gasses can be treated as a monatomic ideal gas. Figure 1 - A cylinder of length L = 1.00 m and inside radius r = 15.0 cm is filled with 5.00 moles of xenon gas. A dise of the same radius and negligible thickness, with a mass of m= 1.75 kg, is slowly lowered into the cylinder, compressing the gas. Then the rest of the cylinder is filled with 3.00 moles of helium gas, and sealed. Both gasses are maintained at a constant temperature of T-100.0°C. (Not to scale.) a) What is the height of the disc, h? b) What is the total internal energy of the xenon gas? (In a monatomic ideal gas, the internal energy is the same as the translational kinetic energy.) c) What is the total internal energy of the helium gas? d) If you could measure the kinetic energy of some arbitrary single molecule of the xenon gas, what would it most likely be? e) If you could measure the kinetic energy of some arbitrary single molecule of the helium gas, what would it most likely be?

Answer 1

Solution :

   Given that

Solution a Given that 15 cm T = 1000 =373.15K Here, - Dine mass s is 1.75 kg Molar mass of He = ug (mole Motar Man of Xe = 131. 29 gimole Mass of Smdes of Xe = 131.29 x5 X - 6SG 4 Man of 3 moles of He gas = 3xys He = 129 Total man of Xe gas = 1.15tq + 10 Loo kg = 1:15 +0.012 = 1.762 kg

force Exerted on the gas = 1.76289.8 = 17.28N the Height of the sinc, h? Any: What Given is that Here, in tegas gas Equation Apply Ideal the NOO., PV = nRT 1. Ah =ART th=nRT = nRT wlow) n = 5 R = 8.315 T =373.15 F -17.28 h 5 X 8'315 X 373.15 14.28 th= 897cm] There fore , h=897cm

13 b. what is the total Internal Energy of the Xenon gas? Ang. Given that Here, The Total Internal energy of Xe gas Where, R = 8.315 T = 373.15 Uxe = 3 / 2 ORT Therefore, ŽXS X 8.315 × 373.15 Tuxe = 2327015 J What is the To ů the Total Internal Energy of the Helium gas? Any:- Given that The Total Internal Energy of He gas - He gas= 3 2nRT There for the = 3 x 3 x 8.315 × 373:15 He gas = 13,962.34 J

d. If you of some could measure the kinetic Energy Arbitary single moleculle of the it most likely be } Xenon gas., What would Ans: Given that Kinetic Energy of Xe molecule where ) K=1.36 1 = 373.15 NOWY Kinetic Energy of Xe molecule = 3 T 3x 138X1023 x373.15 Kinetic Energy of Xe molett +42x10-215 There fore, - kinetic Energy of Xe mole = 1.728 102 / e. If you could measure the kinetic Energy of Some Arbitary single molecule of the Helium gas. What would it most likely be? Any: Given that

Here, kinetic Energy of He molecule elo 388102 x 343.15 He mol = 7.7 2010-21 J There fores He molecule = 7.72x10 a J X-