CAS CH 112 Lecture Notes - Lecture 5: Lattice Energy, Molar Volume, Ideal Gas

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24 Mar 2019

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No expansion work is done when the process occurs at constant volume: u = q. When the system is at constant pressure u = q+w = q p v. The total amount of energy used on a system which includes the energy associated with a change of volume at constant pressure: q= u + p v = h. Accounts for energy in/out when a system undergoes chemical or physical change when pressure is constant. At constant volume cv = u/ t and at constant pressure cp = h/ t: for an ideal gas cp= cv + (p v/ t) = cv + nr where n is moles of ideal gas. If the moles or volume of a gas change, then h doesn"t = u. What is in h: thermal energy from molecular motion, potential energy from bonds and intermolecular interactions, non-expansion work (electrical work) H for a process is negative for the reverse.

Related Questions

Two moles of oxygen gas, which can be regarded as ideal with Cp=29.4 J/(K*mol) (independent of temperature), are initially at 298K in a volume of 12.5 dm^3. The gas is expanded reversibly to 353K at constant pressure. Calculate the final volume and q, w, delta U, and delta H.

Suppose that the gas in the previous question is reversibly compressed to half its volume at constant temperature (298K). Calculate the final pressure and q, w, delta U, and delta H.

An ideal gaseous reaction occurs at a constant pressure of 40.0\({\rm atm}\) and releases 56.7\({\rm kJ}\) of heat. Before the reaction, the volume of the system was 8.80\({\rm L}\) . After the reaction, the volume of the system was 2.40\({\rm L}\) .

Calculate the total internal energy change, \(\texttip{\Delta U}{Delta U}\), in kilojoules.

An ideal gas is confined to a container with a massless piston at the top. (Figure 2) A massless wire is attached to the piston. When an external pressure of 2.00 \(\rm atm\) is applied to the wire, the gas compresses from 4.90 to 2.45\({\rm L}\) . When the external pressure is increased to 2.50 \(\rm atm\), the gas further compresses from 2.45 to 1.96\({\rm L}\) .

In a separate experiment with the same initial conditions, a pressure of 2.50 \(\rm atm\) was applied to the ideal gas, decreasing its volume from 4.90 to 1.96\({\rm L}\) in one step.

If the final temperature was the same for both processes, what is the difference between \(\texttip{q}{q}\) for the two-step process and \(\texttip{q}{q}\) for the one-step process in joules?

Any help is highly appreciated!

Thanks in advanced .

CAS CH 112 Lecture Notes - Lecture 5: Lattice Energy, Molar Volume, Ideal Gas

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