From the given isotherm for one mole of an ideal gas, which follows Boyle's law, what will be the value of temperature? (R = 0.0821 litre atm mol−1K−1).
No. of moles of gas = 1 The gas follows Boyle's law. The given isotherm is between log10 P vs log10 V Now, ideal gas equation is pV = nRT where, p = Pressure V = Volume n = No. of moles R - Gas constant T = Temperature For 1 mole: pV= RT (i) Taking logarithm both sides of equation (i). we get log10pV = log10RT log10p+log10V = log10RT (Since log ab = log a + log b) log10p = - log10V+log10RT Comparing above equation with equation of straight line y = mx + c Thus. Intercept, c = log10 RT Slope, m = -1 Now, from the graph Intercept, c = 2 Thus, log10RT = 2 RT = (10)2 RT = 100 T =