Two moles of a monatomic idea gas are taken through a three process thermodynami

Question

Two moles of a monatomic idea gas are taken through a three process thermodynamic cycle. The gas starts with a pressure PA = 16 atm and a volume VA = 11. The gas in then expanded adiabatically to point B. The temperature at point B is Tn = 25 K. The gas is then compressed isobarically to point C where the volume is equal to VA. The gas is finally returned to state a via a constant volume process. For all calculations using the ideal gas law assume that R = 0.08 L - atm/mol K. For any calculations of work, heat, internal energy, or entropy use R = 8.0 J/mol Compute all unknown pressures, temperatures, and volumes and organize them into a table. Find the work done by the gas going from point A to point B? What is the heat added to the system going from point B to point C? What is the change in internal energy going from point C to point A? What is the change in entropy going from A to B? A ship has a mass of 70.000 kg and is made of a metal that has a density of 6000 kg/m . The ship floats in water (rho = 1000 kg/m ) such that 70 % of its volume is below the surface of the water Assume that the weight of the air is negligible What is the volume of the ship? What is the volume of the empty space inside of the ship? The ship begins to take on water at a rate of 10 L/ . How long will it tale for the ship to sink below the surface of the water? A yo - yo is constructed with two plastic disks and a solid aluminum rod. The disks each have a mass of 80.0 grams with a diameter of 15 0 cm A massless string is wrapped around the rod. One end of the string is held at a constant tension T and the yo - yo is released The rod has a mass of 50.0 grams with a diameter of 5.00 cm. The moment

Explanation / Answer

a. You'll need to first find the force of gravity acting on the ship. F = mg
Fgrav = (70,000)*9.8 = 686000 N

The buoyant force is equal to the amount displaced (water volume) x water density (1000) x g

Fbuyant = Vol*1000*9.8 = 9800*v N

Since the ship is floating, and the buyant force and gravitational force are acting in opposite directions, you can add these forces up (be careful of signs) and set = 0

9800v - 686000 = 0 Thus, the volume displaced (v) = 70m3 (note: this is both the volume of water displaced and the volume of the ship part that is submerged.

Since 70% of the ship's volume is submerged, 70 = .7*totalvolume

Thus, the total volume of the ship = 100 m3

b. If the ship has a volume of 100 m3, a density of 6000 km/m3, and a mass of 70,000 km: We can divide the mass by the density to give us the volume of metal.

vol metal = 70,000/6000 11.6 m3 (note that the "km" units cancell out)

if the total volume is 100 and the metal volume is 11.6, the air volume will be 100 - 11.6 88.4

c. When the water is entering the ship, we can assume that it will be changing it's mass and not necessarily it's volume. The mass entering the ship as a function of time can be modeled by:

"t" (time) * the volume of water entering per unit time [L/sec] * density of water [km/L], which will give us units of [km/sec] which is what we want.

dm/dt = 10*t*1000 = 10000t

the boat starts off with mass 70,000, so the time-dependent forumla for the boat's mass is

M(t) = 70000 + 1000t

We can return to our summation of forces (gravity and buyancy) in part a) to find out at what time the boat sinks (or when the sum of the forces = 0, assuming complete submersion).

F = Fbuyancy + Fgravity = (total volume submerged)*(density water)*(g) - (M(t))*(g) = 0

= 100*1000*9.8 - (70000 + 1000t)*9.8 = 0

980000 = 686000 + 9800t

294000 = 9800t

t = 30 seconds

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