InterviewSolution
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InterviewSolution
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Making use of the result obtained in the foregoing problem. Find the what temperature the isothermal compressibility `x` of a Van der Walls gas is greater than that of an ideal gas. Examine the case when the molar volume is much greater than the parameter `b`. |
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Answer» For an ideal gas `K_0 = (V)/(RT)` Now `K = ((V - b)^2)/(RTV){1 - (2a(V - b)^2)/(RTV^3)}^-1 = K_0(1 - (b)/(V))^2{1 - (2a)/(RTV)(1 - (b)/(V))^2}^-1` =`K_0{1 - (2b)/(V)+(2a)/(RTV)}`, to leading order in `a, b` Now `K gt K_0`if `(2 a)/(RTV) gt (2 b)/(V)` or `T lt (a)/(bR)` If `a,b` do not vary much with temperature, then the effect at high temperature is clearly determined by `b` and its effect is repulsive so compressibility is less. |
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