Problem: Imagine a container placed in a tub of water, as depicted in the accompanying diagram.If neither the volume nor the pressure of the system changes during the process, how is the change in internal energy related to the change in enthalpy?

FREE Expert Solution

Internal Energy:


E=q+w


Recall that work (w) is given by:


w=-PΔV


• (+): heat, q is gained by the system (i.e. endothermic rxn )

• (–): heat, q is released by the system (i.e. exothermic rxn)


• (+): work is done by the surroundings to the system (compression)

• (–): work is done by the system to the surroundings (expansion)


At constant P → q = ΔH (enthalpy)

At constant V → w = 0

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Problem Details

Imagine a container placed in a tub of water, as depicted in the accompanying diagram.

A large cylinder is at 290 Kelvin.  Within it is a smaller cylinder, at 350 Kelvin.

If neither the volume nor the pressure of the system changes during the process, how is the change in internal energy related to the change in enthalpy?

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