Thermodynamics adiabatic transformation I have the following question as howework, I think I don't have enough information to solve it. 
Given are $p_1=311$ kN/m$^2$, $T_1=333$ K, $T_2=853$ K. 
Asked are $p_2$, ratio $V_1/V_2$, and $R$ if $w=372,32$ kJ/kg
How do I go about it? I can use the work equation to get a ratio between $k$ and $R$, but if I plug that into another equation I always remain with two unknowns. 
I can't ask on Physics-meta, I'm sorry. I would really appreciate if someone showed me the way
 A: Although you’ve probably have already found the solution to the problem, here is one approach (though not necessarily the most efficient). It assumes (1) an ideal gas, (2) a closed (constant mass) system (such as a gas in a cylinder fitted with a piston and (3) an adiabatic and reversible (isentropic) process. The work, heat, internal energy, specific heats, and volumes are all lower case to indicate they are per unit mass (specific values). 
We start with the work done in the reversible adiabatic (isentropic) process, given by:
$$ w = \frac{R(T_1 – T_2)}{1-k} \tag{1}$$
Since the initial and final temperatures are given, we have two unknowns, $R$ (specific gas constant, which you need to determine) and $k$.
For an ideal gas, 
$$ k = \frac{c_p}{c_v} \tag{2} $$
where $c_v$ , the specific heat at constant volume, can be calculated as follows. 
From first law:
$$\Delta u = q– w$$
with $q=0$ (adiabatic process) and $\Delta u = – w$, where $w$ is positive if work is done by the gas (expansion) and negative if work is done on the gas (compression).  Since the adiabatic processes increases the temperature ($T_2 > T_1$) the process must be compression, thus $w$ is negative.  
$$w = - 372.32 \text{kJ/kg}$$
$$\Delta u = - (- 372.32 \text{kJ/kg}) = +372.32 \text{kJ/kg}$$
But also, for an ideal gas in a closed system, $\Delta u$ depends only on the temperature change and is given by the following that applies to any process (not just constant volume)
$$\Delta u = c_v \Delta T = c_v (T_2 – T_1),$$
then
$$c_v = \frac{+372.32}{T_2 – T_1}\tag{3}$$
Next we introduce the relationship between specific heats and the specific gas constant for an ideal gas:
$$c_p – c_v = R$$ 
thus 
$$c_p = c_v + R \tag{4}$$
recall (2) $k = c_p / c_v$ and rewrite
$$c_p = k c_v \tag{5}$$
From (4) and (5)
$$k c_v =  c_v + R $$ 
$$k = (c_v + R)/ c_v \tag{6}$$
substitute eq (3)  for $c_v$ in eq (6),  we then have $k$ in terms of $R$, which can be substituted for $k$ in equation (1) to solve for $R$.  After solving $R$ putting the value of $R$ into equation (1) can solve for $k$.
That leaves $v_1/v_2$ and $p_2$ to be determined.  We get these from the equation for a reversible adiabatic process, and the general gas equation for an ideal gas.
Process equation (reversible adiabatic- isentropic):
$$Pv^k = constant$$
$$P_1v_1^k = P_2v_2^k$$
$$\frac{v_1}{v_2} = \left(\frac{P_2}{P_1}\right)^\frac{1}{k} \tag{7}$$
General gas equation:
$$\frac{P_1v_1}{T_1} = \frac{P_2v_2}{T_2}$$
$$\frac{v_1}{v_2} = \frac{P_2T_1}{P_1T_2} \tag{8}$$
Since $k$ has been determined, we have two equations ( (7) and (8) ) and two unknowns  $v_1/v_2$ and $p_2$ which can then be solved.
