Question

The cycle of FIGURE $Ex21.9$consists of three processes. Make a table with rows labeled $A-C$and columns labeled$∆E_{th}$, $W_s$, and $Q$. Fill each box in the table with $+$,$-$, or $0$to indicate whether the quantity increases, decreases, or stays the same during that process.

Short answer

In isothermal process , temperature is kept constant . So change in internal energy is zero . Whatever heat is given to the process , it is converted into work . In adiabatic process , no heat is added nor extracted from the system . So , Q is zero . Work is done by the gas at the expense of internal energy .

Step-by-step solution

Definition of isothermal and adiabatic process

In isothermal process , temperature is kept constant . So change in internal energy is zero . Whatever heat is given to the process , it is converted into work . In adiabatic process , no heat is added nor extracted from the system . So , Q is zero . Work is done by the gas at the expense of internal energy .

Explanation

Process A. The adiabatic expansion process is described by this process. The heat required is $0$In an adiabatic process, there is no heat transfer. and we put $0$for $Q$. The work done$W_s$is positive because the gas is expanding (the volume increases) so we put $+$for $W_s$. From the first law of thermodynamics $Q=W_s+∆E_{th}$. Since $Q=0$then $∆E_{th}=-W_s$i.e. We placed - into that field because the change in thermal energy is negative.

Process B. With rising pressure, this mechanism becomes isochoric. Because the volume does not change in this procedure, the work done is 0, hence we put $0$for $W_s$. The temperature increases since the pressure increases so the change in thermal energy is positive and we put $+$for $E_{th}$. From the first law of thermodynamics $Q=E_{th}+0=E_{th}$so we also put $+$for $Q$.

Process $C$. Isothermal compression is described by this method.. Because $T=const$. in this process, $E_{th}$doesn't change so we put $0$for $∆E_{th}$. We put - in the field according to the job done because the volume decreases. to $W_s$. From the first law of thermodynamics $Q=0+W_s=W_s$so $Q$is negative and we put - into this field.

The table is given below