I had a couple problems on the practice test I was working on, pleas
1. Analyze the pressure vs volume of a bottle engine consisting of 3 liters of an ideal gas as it operates between minimum temperature 300 C and a maximum temperature of 360 C, pumping water to half the maximum possible. Sketch a pressure vs volume graph from the original state to the maximum-temperature state and use the graph to determine the useful work done by the expansion. Then, assuming a diatomic gas, determine the thermal energy required to perform the work and the resulting practical efficiency of the process.
1. PV=nRT where L = 3L, diatomic gas = 5/2 R= 20 J/mol K
I am not sure how to deal with the temperatures, do I just use an average temperature of the min and max? I know how to draw the graph, I think. And in order to find the work I just find the P and then use that in 'dW = P * 'V
Since the Volume is the constant here then we have a Isovolumetric which means that W = 0 and Q = 'dU right?
If the temperature increases from 300 C to 360 C, at constant volume, with water rising in a thin tube, then how high would the water rise?
At what temperature would the pressure be sufficient to support a water column half this high?
If the tube ends at this latter height, then after the system reaches this latter temperature, the gas in the system cannot continue to build pressure and it will begin to expand.
The only way for gas to expand is for water to be displaced from inside the container.
The displaced water will end up coming out the top of the tube, thereby increasing its potential energy.
The amount of water displaced will be equal to the change in the system's volume as its temperature increases (from that latter temperature) to 360 C.
What will be the change in volume?
What will the P vs. V graph look like, as temperature increases from 300 C to the 'half-height' temperature then to the final 360 C temperature?
2. A certain material has density 3.1 kg/L. If 1.86 kg of the material are suspended from a string and immersed in water, what will be the tension in the string?
2. I know that we are using the bouyant force. The Fb = m'g
wt = m *g = 1.86kg * 9.8 m/s^2 = 18.228 kg m/s^2 or N
V = m / rho = 1.86 kg / 3.1 kg/L = .6 L
this is where I get stuck...I know that Tension = Fb * rmom, but I am not sure how to get rmom and since I dont have h either, I dont know where to go.
Good so far, but don't use Fb for m g. m g is the weight of the object, which is the force exerted on it by gravity.
The buoyant force is equal to the weight of the water displaced. The object has volume .6 liter, so it displaces .6 liter of water. What is the weight of that displaced water?
3. There is a small amount of water at the bottom of a sealed container of volume 6.1 L which is otherwise full of an ideal gas. A thin tube open to the atmosphere extends down into the water. The system is initially at atmospheric pressure and temperature 241 C. IF we increase the temperature of the gas until water rises in the tube to a height of 235 cm, then what is the temperature at that instant.
PV = nRT (1 atm)(6.1L) / (514K)(.0821 atm L / mol K) = .144 moles
'dP = rho g 'dh = (1000 kg/m^3)(9.8 m/s^2)(2.35 m)= 23030 Pa = .227 atm
T = (.227 atm)(6.1L) / (.144 mol)(.0821 atm L / mol K ) = 117K
Just wanted to see if I got this one right.
You don't have 6.1 liters of gas at .227 atm.
What you have is 6.1 liters of gas at 1.227 atm.
If you do the same calculation with the correct absolute pressure you'll get the right temperature.