Question

Problem 1 N molecules of carbon dioxide gas Co, (considered as an ideal gas) undergo the cycle shown in the P-V diagram on the right. As a reminder, a CO, molecule has all atoms on a line. Assume that all processes are quasistatic and that the temperature remains such that rotational degrees of freedom are activated, but vibrational degrees of freedom are frozen out. Capital letters A, B, C represent the states at the corners of the cycle, while lowercase letters a, b, c represent the processes between these states. (a) (weight: 1) How many degrees of freedom (both active and frozen out) does a CO, molecule have? How many active ones? achve rotational → 2 degrees as all atoms are in a line for the geometry, can rotate about x and "I Z axis Vibrational 2 degrees per mode of vibration I for Kinetic and I for potential, so here there are 4 degrees frozen out (b) (weight: 1) Assuming one of the processes AB, BC, and CA is adiabatic and another isothermal, identify which is which. Explain your reasoning. An adlabatic process goes from a lower isotherm to a higher isotherm, so I believe it is process CA. Then process BC would be the isotherm. iso therm adiabat 0-0 Q=0

Answer 1

Total No. of degree of freedom for any atonolocale = 3N. where N = No. of atoms. Total degree of freedom for ca = 3x3 active degree of freedoon (linear = 3 Rotational =2) The process Bc is adiabatic and process .ca e isothermal. since. for Isothermal slope of curre=-2 / and for adiabatic slope of curre =- (slope)adea > (slope) gso clearly in picture we can see. cwere BC has greater slope than CAO

(C). ΔΟ + + I (volume a "Wtool cual Cadia) I (VA) (gsotherm for Isothermal process AU=+NRAT = 0. o=00 = NRT ln (N2). adiabatic process for ). foo DU=-W gropari e process (w = out w: