Question

The dot in Fig $\mathbf{19}\mathbf{-}\mathbf{18}\mathit{c}$represents the initial state of a gas, and the adiabatic through the dot divides the p-V diagram into regions $\mathbf{1}$and $\mathbf{2}$. For the following processes, determine whether the corresponding heat is positive, negative, or zero: (a) the gas moves up along the adiabatic, (b) it moves down along the adiabatic, (c) it moves to anywhere in region $\mathbf{1}$, and (d) it moves to anywhere in region $\mathbf{2}$.

Short answer

1. The heat $Q$is when the gas moves up along the adiabatic path.
2. The heat $Q$is when the gas moves down along the adiabatic path.
3. The heat$Q$when the gas moves anywhere in $\mathrm{region}1$is negative.
4. The heat$Q$when the gas moves anywhere in $\mathrm{region}2$is positive.

Step-by-step solution

Stating the given data

The P-V diagram divides into two regions $1$and $2$. The dot represents the initial state of gas at the adiabatic process.

Understanding the concept of heat

The amount of heat Q that is transported to create a temperature change depends on the substance and phase involved, the mass of the system, and the magnitude of the temperature change. We can use the relation between heat and specific heat to find the heat when the gas moves up, down along the adiabatic path, or anywhere in regions 1 and 2.

Formulae:

Energy transferred as heat to the body,$Q=n{C}_p\Delta T$ …(i)

(a) Calculation of the heat when gas moves up the adiabatic path

If the given process is adiabatic, then the total heat is $\mathrm{Q}=0$. For the given condition, the gas moves up at the adiabatic process, which means heat is constant,$\mathrm{Q}=0$.

Hence, the heat when the gas moves up the adiabatic path is zero.

(b) Calculation of the heat when gas moves down the adiabatic path

Similarly, as the gas moves down, the process is adiabatic, so the total heat is $\mathrm{Q}=0$.

Hence, the heat when the gas moves down the adiabatic path is zero.

(c) Calculation of the heat when gas moves anywhere in region 1

Asthegas moves anywhere in regionthe change in temperature is less than zero.

$\begin{array}{c}\Delta T=T_f-T_i\\ <0\end{array}$

Thus, using this above value in equation (i), we get that the heat in region 1 as follows:

$Q<0$i.e., heat is negative.

Hence, the heat when the gas moves anywhere in region 1 is negative.

(d) Calculation of the heat when gas moves anywhere in region 2

As the gas moves anywhere in region $2$, the change in temperature is greater than zero.

$\begin{array}{c}\Delta T=T_f-T_i\\ >0\end{array}$

Thus, using this above value in equation (i), we get that the heat in region 2 is as follows:

$Q>0$i.e., heat is positive.

Hence, the heat when the gas moves anywhere in region 2 is positive.