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Since PV=nRT is used the gas

  1. Mar 7, 2010 #1
    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 isnt 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.
     
    Last edited: Mar 7, 2010
  2. jcsd
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