course Phy 202
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The problem is -- a monatomic gas in a 3.5 liter container is originally at 27 C and atmospheric pressure. It is heated at constant volume until its temperature is 180 C, then at constant pressure until the gas has increased its volume by .52 liters. How much thermal energy is required? By how much does the internal energy of the gas change? How much work is done in the process?
In the first part of the problem (volume is constant) Vo=3.5L, To=300K, Po=1atm and we know that temp changes to 453K. I used PV=nRT to find n, n=1atm*3.5L/[(.0821L*atm/molK)(300K)]=.142mol.
Next I used 'dU=nCv'dT to find Q -- 'dU=.142mol(12.465J/molK)(153K)=271.0J = Q.
For the second part you said to use 'dV and P to find Work--'dV=.52L(.00052m^3) and using PV=nRT I found P for the first part --
P=nRT/V=.142mol(8.21J/mol*K)(453K)/(.0035m^3)=150890.41Pa.
Work=P'dV=.00052m^3*150890.4kg/m*s^2)=78.46J.
Next you said to use the ideal gas law to find the temp required to expand the gas to its new volume-- T=PV/nR=(150890.41kg/m*s^2)(.00402m^3)/(.179mol*8.21J/molK)=412.75K.
You are using .179 mol here, but .142 mol earlier. It's not clear where the 1.79 mol came from, but the system is closed and will have the same number of moles in all states.
If we use the ratio (T2/T1=V2/V1) I get 412.75/453=.92 and .00402/.0035=1.15, this should be equal and is not .92 does not equal 1.15!
I haven't been able to finish the problem because I wasn't sure where I went wrong.
See how it works out with a consistent value of n. This will give you a volume ratio greater than 1.
The next problem is How much ice at 0C should we add to 220 grams of water at 35 C and 16 grams of steam at 100 C to end up with only water at 0C, assuming that all thermal exchanges take place within the s
My solution -- for the first part 220g of water at 35C to 0c I used Q=mc'dt = .22kg(4186J/kgC)(-35C)=-32232.2J, for the second part 16 grams of steam at 100 C to 0C -- Q=.016kg(2010J/kgC)(-100C)=-3216J. You said this was correct but I forgot the energy required to condense the steam. I took the heat of vaporization (540 calories/gram and a calorie is 4.19J, and the heat of fusion for water is 80cal/gram).
540cal*4.19J/1cal=2262.6J/gram, then I took the 220g of water*2262.6J/gram=497772J.
I am still not sure where to go from here. I know that the ice needs to gain energy but I am still a bit confused.
How much energy will all that water and steam lose as they cool to 0 C?
That's how much energy the ice has to give up.
At 80 cal / g = 335 J / g, approx., how many grams of ice would be required?