Solving Entropy and Energy Transfer in a Rigid Tank with Argon Gas

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To determine the change in internal energy of the argon gas in a rigid tank, the heat capacity at constant volume can be used, calculated as (3/2)nR, where n is the number of moles derived from the mass of argon. The change in internal energy equals the heat exchange since no work is done, and the change in entropy for the argon can be calculated using the formula nC_v ln(T_f/T_i). The reservoir's constant temperature allows for the calculation of its entropy change as Q/T, which should be a positive value that exceeds the negative entropy change of the argon. The volume of the tank is not necessary for these calculations, as the ideal gas law can provide it if needed.
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A rigid tank of small mass contains 35.0 g of argon, initially at 240°C and 100 kPa. The tank is placed into a reservoir at 0°C and is allowed to cool to thermal equilibrium?

I don't understand how to 1)determine the change in internal energy of the argon or 2)the energy transferred by heat.

The question asks to find the change in entropy as well the change in entropy of constant-temperature bath.

I believe I have to start with finding the volume of the tank but am not sure. Anyone want to help
 
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Hint

If I recall, the heat capacity of a monatomic gas at constant volume is

(3/2) n R.

So you will probably want to calculate the difference in the temperatures involved, and the number of moles (which you can get from mass given to you as data in the problem), and then use that the idea that change in internal energy is heat capacity times change in temperature, when no work is being done.

Since no work is being done, wouldn't the heat exchange equal the change in internal energy? (Overlooking any difference in signs, of course.)
 
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Check me on this, but at constant volume, is the change in entropy of an ideal gas

nC_v ln(T_f/T_i)

where n is number of moles, C_v specific heat at constant volume [which I am thinking is (3/2)R], T_f final absolute temperature, T_i initial absolute temperature?

If I am right about that, then you can get the entropy change in the argon. It had better come out negative!

The reservoir is assumed to be at constant temperature, and you by now will know the heat Q transferred to it, as well as its temperature. So you can easily calculate the entropy change in it as Q/T. This will be a positive number, and its magnitude has to be at least as much as the magnitude of the drop in entropy of the argon, or else something is wrong with your world.
 
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If I am right, you never need to know the volume of the argon tank. If you really wanted to, you could figure it out from the initial conditions after you calculate the number of moles based on the mass of the argon, using the ideal gas law. Argon, being an inert gas, is monatomic.
 
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