A mole of a monoatomic ideal gas at 1 atm and 273 K is allowed to expand adiabatically against a constant pressure of 0.395 bar will until equilibrium is reached. (a) What is the final temperature ? (b) What is the final volume ? (c ) How much work is done by the gas ? (d) What is the change in internal energy ?

A mole of a monoatomic ideal gas at 1 atm and 273 K is allowed to expa

1.	A mole of a monoatomic ideal gas at 1 atm and 273 K is allowed to expand adiabatically against a constant pressure of 0.395 bar will until equilibrium is reached. (a) What is the final temperature ? (b) What is the final volume ? (c ) How much work is done by the gas ? (d) What is the change in internal energy ?
Answer» Solution :LET the initial and FINAL volumes of the gas be `V_(1)` and `V_(2)m^(3)` respectively. Given that the initialtemperature is 273K. Let the final temperature be `T_(2)` We have, `p_(1)V_(1)=n_(1)RT_(1)` `V_(1)=(1xx8.314xx273)/(1xx10^(5))=0.022697m^(3)` For an ADIABATIC expansion of 1 mole of monoatomic ideal gas against a constant external pressure `(p_(2))`, work done is given as `W=-p_(2)(V_(2)-V_(1))=C_(V)(T_(2)-T_(1))=(3R)/(2)(T_(2)-T_(1))` or `-0.395xx10^(5)(V_(2)-0.022697)=(3xx8.314)/(2)(T_(2)-273)`............(1) Again, `p_(2)V_(2)=nRT_(2)` `0.395xx10^(5)xxV_(2)=1xx8.314xxT_(2)` ..............(2) Solving eqns. (1) and (2) , we get (a) the final temperature, `T_(2)=207K` (b) the final VOLUME `V_(2)=0.043578m^(3)` (c ) the work done by the gas `W=-p_(ext)(V_(2)-V_(1))` `=-0.395xx10^(5)(0.043578-0.022697)` `=-825J//mole` (d) as `q=0`, and `q=DeltalI-W` `DeltalI=W=-825J//mole`