1. An ideal gas undergoes a reversible expansion at 25.0 o C . During the expans
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Question
1. An ideal gas undergoes a reversible expansion at
25.0 oC. During the expansion, the gas does 1825 J of work moving the piston upward as indicated in schematic below.
(a) What is the change in entropy ?S of the gas?
(b) Suppose the volume of the gas is doubled during the expansion. What is the number of moles n of the gas?
2. A uniform horizontal beam has a mass of 13.0 kg . As you see below, the beam makes an angle of 45 degrees with a vertical pole. The beam
An ideal gas undergoes a reversible expansion at 25.0 degreeC. During the expansion, the gas does 1825 J of work moving the piston upward as indicated in schematic below. (a) What is the change in entropy ?S of the gas? (b) Suppose the volume of the gas is doubled during the expansion. What is the number of moles n of the gas? A uniform horizontal beam has a mass of 13.0 kg. As you see below, the beam makes an angle of 45 degrees with a vertical pole. The beam's right end is connected to a horizontal, light (essentially mass-less ) cable. The beam's left end is hinged at pivot A on the rigid, fixed pole. The left end of the cable is also attached to the pole. See figure. Show all work. (a) What is the tension T in the cable? (b) What is the horizontal component of force FH exerted by pivot A on the beam? (c) What is the vertical component of force Fv exerted by pivot A on the beam ? (d) Assuming the beam length is 1.0 m, suppose a vandal severs the cable such that the tension becomes zero. Consider the moment just after the cable is cut and the beam is in a horizontal position. At that instant, what is the magnitude of the torque (about pivot A) exerted by the gravitational force? What is the magnitude of the beam's angular acceleration about point A at the instant?Explanation / Answer
1)
a) dS = dQ/T = dW/T = (1825)/(273.15+25) = 6.1211 J./K
b)dW = n*R*T*logV2/V1)
1825 = n*8.314*(273.15+25)*log(2V/V)=====> n = 2.4457 moles
2)
a) T*L*sin45 = M*g*(L/2)*cos45
T = M*g/2 = 63.765 N
b) Fh-T = 0
Fh = T = 63.765 N
Fv - Mg = 0
Fh = Mg = 127.53 N
d) Torque = M*g*L/2 = (13*9.81*1)/2 = 63.765 N
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