Work done in isothermal process,

W = $$-$$ nRT ln$${{{V_2}} \over {{V_1}}}$$

$$ \therefore $$   $$\left| W \right|$$ = nRT ln$${{{V_2}} \over {{V_1}}}$$

$$ \Rightarrow $$   $$\left| W \right|$$ = nRT (lnV₂ $$-$$ lnV₁)

$$ \Rightarrow $$   $$\left| W \right|$$ = nRT lnV₂ $$-$$ nRT lnV₁

given that $$V$$₂ $$=$$ $$V$$

$$ \therefore $$   $$\left| W \right|$$ = nRT lnV $$-$$ nRT lnV₁

By comparing with the straight line equation,

y = mx + C

We get slope is nRT and intercept $$-$$ nRT lnV₁ in $$\left| W \right|$$ and lnV graph.

As T₂ > T₁   So,

Slope nRT₂ > nRT₁

and intercept

$$-$$ nRT₂ lnV₁ < $$-$$ nRT₁ lnV₁

So, we can say

(1)   slope of T₂ line is more then T₁

(2)  intercept of T₁ line is less negative than T₂ line, and intercept of T₁ can't be positive, it can be 0 or less than 0 as $$-$$ nRT₁lnV₁ always $$ \le $$ 0