WTST

Hi, I'm having some problems with this exercise on thermodynamics: "A cylinder with a movable piston contains a certain quantity of helium. With a very slow transformation - represented in a V-p graph as a straight line - it goes from state [latex]A(p_A=40kPa, V_A=3dm^3, T_A=300K)[/latex] to state [latex]B(p_B=150kPa, V_B=1dm^3)[/latex]. It is then subjected to a costant-volume transformation followed by a costant-pressure one, returning to the initial state". It can be summarized in this graph: all From which it's pretty simple to find the work done (area of the triangle): [latex](110kPa*2dm^3)/2=(110Pa*2m^3)/2 \Rightarrow \color{red}{W=-110J}[/latex]. A clock-wise transformation means positive work, but this graph - misdirected by the text of the exercise - is all the opposite (V-p, instead of p-V), hence the negative work. It then asks for [latex]T_B[/latex], which I calculated from the ideal gas law. I didn't know the amount of substance, but I took that from state A. [latex]n=\frac{p_AV_A}{RT_A} \Rightarrow \color{red}{n=0,048}[/latex] then [latex]T_B=\frac{p_BV_B}{nR} \Rightarrow \color{red}{T_B \approx 375K}[/latex]. Finally, it wants the heat transfer with the outside during [latex]A\rightarrow B[/latex]. At first, I wanted to use [latex]Q=cm\Delta T[/latex], but the specific heat gave me some doubts: which one should I use when the transformation is not particular (I'm referring to the fact that there are specifc specific heats when the transformation is with costant pressure/costant volume)? Then I thought I could use [latex]\Delta U=Q-W[/latex]. I could calculate the difference in internal energy: [latex]\Delta U=\frac{3}{2}NK_b \Delta T \Rightarrow \Delta U=\frac{3}{2}(n*N_A)K_b \Delta T \Rightarrow \Delta U=\frac{3}{2}*2,89^{22}*1,38^{-23}*75 \Rightarrow \color{red}{\Delta U_{A\rightarrow B} \approx +45J}[/latex]. Since it is work done on the gas, it is more internal energy, not less: [latex]\Delta U=45=Q \color{red}{+} W[/latex]. Also, we are only talking about the work done from A to B, which includes the area underlying the triangle: [latex]\color{red}{W_{A\rightarrow B}=220J}[/latex]. But I can't figure out the sign of the Q: the work causes a compression, but I don't know how much. I mean: it could be so much that the temperature went much higher than 375K, arriving then at that level giving off heat; but it could also be that the work brought the temperature under 375K and that it arrived at that level absorbing heat. And anyway, whatever sign I use the answer [latex](Q=-145J[/latex], the book says) isn't right. So, where's my mistake? Thanks in advance. (If something isn't clear, please tell me)