CHEMISTRY COMMUNITY

During the test of an internal combustion engine, 3.00 L of nitrogen gas at 18.5 C was compressed suddenly (and irreversibly) to 0.500 L by driving in a piston. In the process, the temperature of the gas increased to 28.1 C. Assume ideal behavior. What is the change in entropy of the gas?


Anybody help me out with thus one? Do you have to do 2 separate calculations and add them together to get the total dS?

Yes, you do 2 separate change in entropy equations (one is decreasing from the decrease in vol and the other is increasing from the increase in temp). Finally,you add them up for the net entropy change.

Why can't you use the heat capacity in calculating the change in entropy for the temperature increase? Wouldn't the heat capacity be (5/2)R because it is an ideal gas? The solutions manual only uses nRln(T2/T1) and I thought it would make sense to use nCln(T2/T1).

Yeah, I'm also confused as to why the solution manual uses delta s= nRln(T2/T1) rather than delta s=nCln(T2/T1) when the temperature change occured.

For this question the amount of moles is not given. Why does the calculation in the solutions manual say 1.00 mol? Is it because we are doing the calculation with a gas that exhibits "ideal gas" behavior?
Thank you in advance for your help

csebastiani_1B wrote:During the test of an internal combustion engine, 3.00 L of nitrogen gas at 18.5 C was compressed suddenly (and irreversibly) to 0.500 L by driving in a piston. In the process, the temperature of the gas increased to 28.1 C. Assume ideal behavior. What is the change in entropy of the gas?


Anybody help me out with thus one? Do you have to do 2 separate calculations and add them together to get the total dS?

Yes, you have to do 2 separate the calculations because the s=nR ln V2/V1 can be used only when it is isothermal and reversible. And S= n Cv ln T2/T1 is used when volume is constant.