Simple thermodynamics basics - can internal energy be calculated? Well this got me stumbled, because I've been wondering what the "question" is. One of the example examns got the following question (strangely there are no supplied solution books):

A $0.2 {\rm m^3}$  thermally insulated rigid container is divided into two
equal volumes by only a thin membrane. Initially, one of these
chambers is filled with air at a pressure of $700\, {\rm kPa}$ and $37 \, {\rm C}$ while
the other chamber is evacuated.
$C_p = 1.005 \frac{{\rm kJ}}{{\rm kg \cdot K}}\,$ and $\, C_v =0.721 \frac{{\rm kJ}}{{\rm kg \cdot K}}$

Now the questions:

a)    Determine the change in internal energy of the air when the membrane is ruptured.
b)    Determine the final air pressure in the container

Is this now a really silly question or am I missing something important? Cause isn't the internal energy an intrinsic property that has to be looked up/ experimentally determined?
And for the second problem, as there can be no energy transfer the internal energy also has to stay the same - so the temperature doesn't change and the pressure simply halves. ($PV = {\rm constant}$). Or am I missing something?
 A: The internal energy of an ideal gas can only change if heat is added or removed from the system, or if the system does some work. Neither is the case in this example so the change in internal energy is zero.
The final pressure is, as you say, just half the initial pressure.
I'm not sure why they give you the specific heats, but you can use them to work out how ideal your gas is. The specific heats of an ideal diatomic gas should be 2.5 and 3.5 joules/mole/R, and converting the specific heats you're given to J/mole/R (assuming an average $M_w$ of 28.8 for air) gives 2.50 and 3.48 to two SF. So approximating the air as an ideal gas seems entirely justified.
Response to comment:
You're given $C_v$ = 721 J/kg/K. The average molecular weight of air is 28.8 (assuming 20% oxygen and 80% nitrogen) so one mole is 0.0288kg. Multiplying by 0.0288 to convert to moles gives $C_v$ = 20.76 J/mole/K. Now divide by the molar gas constant, $R$ = 8.314 J/mole/K, to get $C_v$ = 2.50 (to 2 significant figures). Exactly the same calculation gives 3.48 for $C_p$.
A: The way the question is worded "when the membrane is ruptured" sounds like a Joule expansion.  The answer depends (sort of) on what class this is for.  Based on the title of your question, it sounds to me that this exam is for just a basic thermodynamics course in which case this is likely just a "trick" question.  
So it's just a classic Joule expansion, no change in internal energy, and the pressure is halved (to 350 kPA).  Neither heat capacity is relevant because the process is not reversible, not at equilibrium, and neither the volume nor pressure is constant.  All that added information is just to try to throw you off, to see if you really understand what's going on.  That's my take on this (unless I'm missing something). HTH.
A: This is a case of adiabatic expansion. So change in internal energy is work done by the gas or system.
$dU=-dW=-\dfrac{P_iV_i-P_fV_f}{\gamma -1}\tag*{}$
where $\gamma =C_p/C_v$
Also, change in internal energy is given by,
$\Delta U=mC_v(T_f-T_i)\tag*{}$
where, $T_f=T_i\left(\dfrac{V_i}{V_f}\right)^{\gamma -1}$ and, $m=\dfrac{PV}{RT}M$
$dU=-42,650$ joules
Now final air pressure can be given by,
$P_iV_i^\gamma =P_fV_f^\gamma \implies P_f=266.4$ kPa
Above many answers and comments considered it as irreversible adaibatic case, but this is not. Instead of membrane if piston is placed and which moves quickly upto double the volume of the container. Now piston can be pushed back to same volume and if heat is not exchanged then it can gain its initial state.
The work done on the system to move back to its initial position is equal to work done by the system or gas during expansion. Thus by calculating external work, we can find the change in internal energy during expansion.
Edit: This is to show that there is change in internal energy in OP's question. Calculation from online calculator is pasted.

TWO DOWNVOTES FOR CORRECT ANSWER, NOT FIRST TIME.NOW THREE
