Is Delta Enthalpy Zero for an Isothermal Gas Expansion Process?

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In an isothermal process involving gas expansion, the relationship between delta enthalpy (delta H) and other thermodynamic variables is clarified. While delta H is expressed as delta U plus the work done, it is important to note that delta U is zero in this scenario. The work done is calculated as nC ln(V2/V1), leading to the conclusion that delta H is not zero. However, the correct formulation states that delta H equals delta U plus delta(PV). For an ideal gas, delta(PV) can be expressed as R delta(nT), which equals zero if the number of gas molecules (n) and temperature (T) remain constant. Thus, under these conditions, delta H is indeed zero, contradicting initial assumptions.
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Is delta enthalpy zero for...

Is delta H, for an isothermal process in which a gas expands?
delta H=delta U+work done,correct?
delta U is 0,but work done = nCln(V2/V1).
So,delta H is not equal to 0.But my book says so.
 
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sachin123 said:
delta H=delta U+work done,correct?
.

No. deltaH = deltaU + delta(PV). Work done is p delta(V) (under conditions of constant pressure).

For an ideal gas delta(PV) = R delta(nT). As long as the number of gas molecules in the system (n) and the temperature (T) remain constant, delta(PV) = 0.
 
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