A cylinder contains 0.0100 mol of helium at T = 27.0o C. (a) How much heat is needed to raise the temperature to 67.0o C while keeping the volume constant? Draw a pV-diagram for this process. (b) If instead the pressure of the helium is kept constant, how much heat is needed to raise the temperature from 27.0o C to 67.0o C? Draw a pV-diagram for this process. (c) What accounts for the difference between your answers to parts (a) and (b)? In which case is more heat required? What becomes of the additional heat? (d) If the gas is ideal, what is the change in its internal energy in part (a)? In part (b)? How do the two answers compare? Why?

Solution 20E Introduction The first process in the given problem is isochoric and the second process is isobaric. So we will use the molar specific heat at constant volume (for isochoric process) and at constant pressure (for isobaric process) to calculate the required heat. We have to find the reason why there is a difference in the required heat in those two processes. Then we have to find out the internal energy in both cases and compare the results. (a) Step 1 The heat required to change temperature in an isochoric process is given by Here n is the mol, CV is the molar specific heat at constant volum2,is the final temperature and1T is the initial temperature. We know n = 0.0100 mol C V = 12.47 J/mol.K T = 67°C = 340 K 2 T 1 27°C = 300 K Using this values we have Step 2 This is an isochoric process and hence the graph will be parallel to the pressure axis and since the temperature increases here the pressure will also increase. So the line in PV diagram will go from lower pressure to higher pressure. The diagram is shown below. (b) Step 3 If, instead of volume, the pressure of the gas remains constant then the process will become an isobaric process and we will us the molar specific heat at constant volume to calculate the heat supplied to the system. The molar specific heat of the 2 at constant pressure is given byPC= 20.78 J/mol.K. Hence the energy supplied to increase the temperature is