Answer to Question #141407 in Molecular Physics | Thermodynamics for matome

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

The total work:

$A=A_{AB}+A_{BC}$

where $A_{AB}$ is work that done in isobaric process ($P=const$) , $A_{AB}=P_1 (V_2-V_1)$

and $A_{BC}$ is done in isochoric process ($V=const$). $A_{BC}=0$

So, $A=A_{AB}=P_1(V_2-V_1)$

$A=10000*10^{-3}=10$ J