91Ó°ÊÓ

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 $Q_1=5.0p_1{v}_1$.

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,

$\mathrm{Δ}E=Q−W$ …(¾±)

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

$\mathrm{Δ}{E}_1=5.0p_1{v}_1−W_1$ …(¾±¾±)

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

$\mathrm{Δ}{E}_2=5.5p_1{v}_1−W_2$ (iii)

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

$\mathrm{Δ}{E}_1=\mathrm{Δ}{E}_2$

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

$A=0.5p_1{v}_1$ …(¾±±¹)

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 $1.5$.