Consider The Reversible Isothermal Expansion Of An Ideal Gas In

[latex] w = -nRT ln\frac{V_{2}}{V_{1}} [/latex]

 

[latex] w = -nRT ln\frac{V_{b}}{V_{i}} [/latex]

 

[latex] \left | w \right | = nRT ln\frac{V_{b}}{V_{i}} [/latex]

 

[latex] \left | w \right | = nRT ln V_{b} – nRT ln V_{i} [/latex]

 

[latex] \Rightarrow y = m x – C [/latex]

 

Therefore, the slope of curve 2 is more than curve 1 and the intercept of curve 2 is more negative than curve 1.

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