1. The problem statement, all variables and given/known data A rigid vessel contains 0.1kg of air initially at 1bar and 20degC. A heat transfer to the gas increases the temperature to 1000degC. What type of process does the gas undergo ? Calculate (i) the initial and final volume of the air (ii) the final pressure (iii) the change in internal energy and enthalpy (iv) the work and heat transfers 2. Relevant equations 1st law eqn 3. The attempt at a solution i) pV = nRT using T = 293.15K and finding n using the molar air mass, p is given as 1 bar. solution is around 0.084 m^3 (2sf) , (V = nRT/p) ii) constant vol, so p/T = const. using the initial and final states, finding the 2nd pressure results from some simple calculations i.e p(2) = p(1)/T(1) * T(2). solution is aorund 4.4bar. so far so good. iii) this is where the books' answer and my own are different. My solution: change in internal energy = mass * specific heat at constant vol * temp change mass = 0.1kg temp change = 980 SH =1.5R where R = univ. gas constant / molar mass(air) = 8.3145/0.02897 giving SH = 430.5 (4sf) hence change in internal energy = 42 kJ(2sf) (multiplying the three terms) The book gives 70.6kJ? for the enthalpy, continuing with my solution, it is = change in internal energy + any work done but no work is done, so the enthalpy = the heat transfer is the change in internal energy (const.volume) hence equals 42kJ. The book gives 99kJ? iv) i said no work done, so did book. the book said the heat transfer is 70.6kJ? i would have said my 42kJ. also, despite the question being in the 1st law part, i thought that only applied at constant pressure? so can't i apply it here? i appreciate im asking a lot, any help appreciated. Regards, James.