Answer to Question 1

(a) Use pV = nRT

At A T₁ = p₁V₁/nR = (1 x 10⁵ x 1 x 10^-3)/nR = 100/nR

At B T₂ = p₂V₂/nR = (5 x 10⁵ x 0.2 x 10^-3)/nR = 100/nR

therefore the temperature are the same, so the line must be an isothermal.

 

(b) A to B (heat must be removed to keep temperature the same)

    C to A as the pressure falls in a constant volume, and no work is done.

 

(c) W = pDV = 5 x 10⁵ Pa x (1 x 10^-3 m³ x 0.2 x 10^-3 m³) = 5 x 10⁵ x 0.8 x 10^-3 m³ = 400 J

 

(d) (i)  The temperature is the highest at point C. (This is because if n remains the same and R is a constant, T must be high as P and V are high)

    (ii) T = pV/nR = (5 x 10⁵ Pa x 1 x 10^-3 m³) ÷ (8.3 J mol^-1 x 0.069 mol)

        T = 870 K