Question

calculate the heat and work involved when 2L of ideal gas at 10 atm expands isothermicaly aganist a constant external pressure of 1 atm

Answer 1

I hope you understand !!!
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In the case of expansion of any gas, work is done by the system.

Now we know that 
     w=−pextΔV=−pext(Vf−Vi)

where pext​ = external pressure of the system = 0
(The given pressure of 10 atm in the question, is the pressure of the gas itself. But we need external pressure in the expression. The gas is expanding into vacuum which has no pressure. Thus external pressure = 0)
         Vf = 10 L
          Vi = 2 L

Thus work done by the system will be:
  w = -0×(10-2)
     = 0
Thus no work is done.
Also, heat absorbed (q) is a consequence of gaining temperature, but the system is working at constant temperature (isothermal condition).
q = -w
So there will be no heat absorbed.

Now for part (1)
pext = 1 atm
Vf = 10 L
Vi = 2 L
So w = -1 atm ×(10-2) L = - 8 atm L
Also, during isothermal expansion, 
q = - w = 8 atm L

For part (2)
Now the work is done reversibly. Then the expression of work done will be:
wrev = −∫VfVipindV
Here 'pin' is the pressure  of the gas, and not external pressure.
Since the gas is ideal, so pV = nRT
or pin=nRTV
and,
w=−∫VfVinRTdVV    = −nRT lnVfVi    = −2.303 nRT logVfVi
So, pin = 10 atm = nRTV (V already included in log)
     Vf = 10 L
     Vi = 2 L
So, 
w=−2.303 nRT logVfVi
 = −2.303×10×log102
 
 = −16.1 atm L

Now q = -w = 16.1 atm L