200 J of heat is provided to a gas in a piston which expands against a constant atmospheric pressure. Overall the gas performs 40 J of pressure-volume work. Compute (a) the interna... 200 J of heat is provided to a gas in a piston which expands against a constant atmospheric pressure. Overall the gas performs 40 J of pressure-volume work. Compute (a) the internal energy change ΔE and (b) the enthalpy change ΔH of the gas.
Understand the Problem
The question is asking to compute the change in internal energy (ΔE) and the change in enthalpy (ΔH) for a gas in a piston, given that 200 J of heat is added and 40 J of work is performed by the gas against atmospheric pressure.
Answer
(a) $\Delta E = 160 \, \text{J}$; (b) $\Delta H = 200 \, \text{J}$
Answer for screen readers
(a) $\Delta E = 160 , \text{J}$
(b) $\Delta H = 200 , \text{J}$
Steps to Solve
- Calculate the change in internal energy, ΔE
Using the first law of thermodynamics: $$ \Delta E = Q - W $$
Where:
- $Q = 200 , \text{J}$ (heat added to the system)
- $W = 40 , \text{J}$ (work done by the gas)
Substituting the values: $$ \Delta E = 200 , \text{J} - 40 , \text{J} $$
- Evaluate the equation for ΔE
Now perform the calculation: $$ \Delta E = 160 , \text{J} $$
- Calculate the change in enthalpy, ΔH
Enthalpy change for processes occurring at constant pressure can be given by: $$ \Delta H = Q $$
Since the heat added at constant pressure is directly used for enthalpy: $$ \Delta H = 200 , \text{J} $$
(a) $\Delta E = 160 , \text{J}$
(b) $\Delta H = 200 , \text{J}$
More Information
- The internal energy change reflects the energy change within the system due to the heat added and work performed.
- The enthalpy change at constant pressure is equal to the heat added to the system.
Tips
- Confusing work done on the system versus work done by the system. In this case, the gas performs work, so it is subtracted.
- Forgetting that the enthalpy change at constant pressure is equal to the heat added (not including work).
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