Can Two Points on a PV Curve Represent the Same State in Thermodynamics?

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suppose an Q amount of heat is given to a system initially at v_1 & p_1 vol and pressure. and the system does a equal amt of work on the surroundings so that delU=0. but in doing this work system has expanded to volume v_2 and has a pressure p_2 due to which it is at a different position on the P-V curve. is it possible that two points on the PV curve represent the same state.

Is free expansion of a gas a reversible process if so how p_ext in this case is always far lower than p_int and the system is never in equilibrium through out the process. how can be then free expansion achieved reversibly. (for the process to be reversible p_ext shuld be infinitesimal greater than p_int isn't it so?)


In case of enthalpy delH=delU + deln_g(RT). this relation is derived assuming that the process is isothermal. and since PV=nRT is used the gas is assumed to be ideal then why in this case is delU is not taken as zero? where n_g is the difference between the no of moles of product formed and the moles on the reactant side.
 
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initiator said:
suppose an Q amount of heat is given to a system initially at v_1 & p_1 vol and pressure. and the system does a equal amt of work on the surroundings so that delU=0. but in doing this work system has expanded to volume v_2 and has a pressure p_2 due to which it is at a different position on the P-V curve. is it possible that two points on the PV curve represent the same state.
Of course, for isothermal expansion of an ideal gas, the final state can be the same as the initial state if no expansion has occurred.
Is free expansion of a gas a reversible process if so how p_ext in this case is always far lower than p_int and the system is never in equilibrium through out the process. how can be then free expansion achieved reversibly. (for the process to be reversible p_ext shuld be infinitesimal greater than p_int isn't it so?)
Free expansion is always irreversible. For expansion to be reversible, the expansion must be quasi static such that the external pressure is only slightly less than the mean pressure of the gas.
In case of enthalpy delH=delU + deln_g(RT). this relation is derived assuming that the process is isothermal. and since PV=nRT is used the gas is assumed to be ideal then why in this case is delU is not taken as zero? where n_g is the difference between the no of moles of product formed and the moles on the reactant side.
The internal energy of an ideal gas reaction mixture is not a function only of temperature. It changes as a result of making and breaking chemical bonds, so that there are changes in the amounts of the various chemical species that are present. So, when a chemical reaction occurs at constant temperature, ##\Delta U## is not equal to zero.