Question

The p-Vdiagram in Fig. 18-60 shows two paths along which a sample of gas can be taken from state ato state b, where ${\mathit{v}}_{\mathbf{b}}\mathbf{=}\mathbf{}\mathbf{3}\mathbf{.0}{\mathit{v}}_{\mathbf{1}}$. Path 1 requires that energy equal to$\mathbf{5}\mathbf{.0}{\mathit{p}}_{\mathbf{1}}{\mathit{v}}_{\mathbf{1}}$ be transferred to the gas as heat. Path 2 requires that energy equal to $\mathbf{5}\mathbf{.5}{\mathit{p}}_{\mathbf{1}}{\mathit{v}}_{\mathbf{1}}$ be transferred to the gas as heat.What is the ratio ${\mathit{p}}_{\mathbf{2}}\mathbf{/}{\mathit{p}}_{\mathbf{1}}$?

Short answer

The ratio $\frac{p_2}{p_1}$is 1.5.

Step-by-step solution

Stating the given data

1. The volume at point b is $v_b=3.0v_1$
2. The energy required for path 1 is$Q_1=5.0p_1{v}_1$
3. The energy required for path 2 is $Q_1=5.0p_1{v}_1$.

Understanding the concept of first law of thermodynamics

Energy can only be converted, not created or destroyed, according to the first law of thermodynamics. We can use the concept of the first law of thermodynamics to get the required value of the ratio of pressures. These pressures are found using the formula of work done by the body.

Formula:

Internal energy of a body, using the first law of thermodynamics,

$\Delta E=Q-W$ …(i)

Calculation of the required ratio of pressures

Using equation (i), the value of internal energy for path 1 is given by

$\Delta E_1=5.0p_1{v}_1-W_1$ …(ii)

Using the same equation (i), the value of internal energy for path 2 is given by

$\Delta E_2=5.5p_1{v}_1-W_2$ (iii)

The change of internal energy during the process remains constant;hence

$\Delta E_1=\Delta E_2$

Using equations (ii) and (iii), we get the work difference as

$W_2-W_1=0.5p_1{v}_1$

$W_2-W_1$can be written as the area of the triangle(A) using the work-graph. Thus, the above equation is given by

$A=0.5p_1{v}_1$ …(iv)

The area inside the triangle can be written as

$A=\frac{1}{2}(v_b-v_1)(p_2-p_1)$ (v)

Since ,$v_b=3v_1$ from equation (v) we get the area value as

$\begin{array}{c}A=\frac{1}{2}(3v_1-v_1)(p_2-p_1)\\ =\frac{1}{2}\left(2v_1\right)(p_2-p_1)\\ =v_1(p_2-p_1)\\ 0.5p_1{v}_1=v_1(p_2-p_1)\text{(fromequation(iv))}\\ 0.5=\frac{1}{p_1}(p_2-p_1)\\ 0.5=\frac{p_2}{p_1}-1\\ \frac{p_2}{p_1}=1.5\end{array}$

Hence, the required ratio is $1.5$.