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A 22.0 L volume of an ideal gas in a cylinder with a piston is at apressure of 3.0 atm. Enough weight is suddenly removed from thepiston to lower the external pressure to 1.5 atm. The gas thenexpands at constant temperature until its pressure is 1.5atm.

Find ?E for this change in state.
Find ?H for this change in state.
Find q for this change in state.
Find w for this change in state.

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Jean Keeling
Jean KeelingLv2
28 Sep 2019

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Related questions

1. Find deltaE for this change in state
2. Find deltaH for this change in state
3. Find q for this change in State
4. Find w for this change in State

1. An ideal gas sealed in a rigid 4.04-L cylinder, initially at pressure Pi=12.00 atm, is cooled until the pressure in the cylinder is Pf=1.39 atm. What is the enthalpy change for this process?

2. What is the change in internal energy for this process?

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?



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