Explain the isothermal and free expansion of an ideal gas

1.	Explain the isothermal and free expansion of an ideal gas
Answer» <P> SOLUTION :For ISOTHERMAL (T = CONSTANT) expansion of an ideal gas into vacuum, `w=0` since `p_(ex) = 0`. Also, Joule determined experimentally that `q=0`, THEREFORE, `Delta U=0` `Delta U= q+w` can be expressed for isothermal irreversible and reversible changes as follows: 1. For isothermal irreversible change `q=- w = p_(ex) (V_(f) - V_(i) )` 2. For isothermal reversible change `q=-w = nRT "ln" (V_f)/( V_i) = -2.303 nRT log "" (V_f)/( V_i)` 3. For adiabatic change, `q=0` `sigma Delta U= W_("ad")`

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