Answer to Question #139690 in Molecular Physics | Thermodynamics for Kgokane Sekgoka

Question

A monatomic ideal gas at 27°C undergoes an isobaric process from state A to B, followed by an isochoric process from state B to C. What is the total work done by the gas in these two processes? (A = 2 atm, 1 L; B = 2 atm, 2 L; C = 1 atm, 2 L)

Answer 1

$T = 27\degree{C} = 300K$

$A = 2atm, \ 1L\ (i.e \ p_1 =2\,atm,\ V_1 =1L )\\ B = 2atm, \ 2L\ (i.e \ p_2 =2\,atm,\ V_2 =2L )\\ C = 1atm, \ 2L\ (i.e \ p_3 =1\,atm,\ V_1 =2L )$

$Process_{A\to B}=Isobaric\\ Process_{B\to C}=Isochoric$

$Work\ during\ an\ isobaric\ process=p(V_2-V_1)\\ W_{A\to B} = p(V_2-V_1)\\ W_{A\to B} = 2(2-1)\\ W_{A\to B} = 2(1)\\ W_{A\to B} = 2\,atmL$

$Work\ during\ an\ isochoric\ process=0J\\ W_{B\to C} = 0J = 0\,atmL$

(This is because the volume is constant, hence no work is done on the surroundings)

$\begin{aligned} Total\ Work\ Done& = W_{A\to B} + W_{B\to C}\\ &= 2\,atmL + 0\,atmL\\ & = 2\,atmL\\ &= 2(101325Nm^2/atm)(10^{-3}m^3/L)atmL\\ &= 202.65J \end{aligned}$

$\therefore$ The Total Work-done by the gas in the two processes is $202.65J$