InterviewSolution
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InterviewSolution
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One mole of a certain gas is contained in a vessel of volume `V = 0.250 1`. At a temperature `T_1 = 300 K` the gas pressure is `p_1 = 90 atm`, and at a temperature `T_2 = 350 K` the pressure is `p_2 = 110 atm`. Find the Van der Walls parameters for this gas |
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Answer» `p_1 = RT_1 (1)/(V^2), p_2 = R T_2 (1)/(V - b) - (a)/(V^2)` So, `p_2 - p_1 = (R(T_2 - T_1))/(V - b)` or, `V - b = (R(T_2 - T_1))/(p_2 - p_1)` or , `b = V - (R(T_2 - T_1))/(p_2 - p_1)` Also, `p_1 = T_1 (p_2 - p_1)/(T_2 - T_1) -(a)/(V^2)`, `(a)/(V^2) = (T_1(p_1 - p_2))/(T_2 - T_1) - p_1 = (T_1 p_2 - p_1 T_2)/(T_2 - T_1)` or, `a = V^2 (T_1 p_2 - p_1T_2)/(T_2 - T_1)` Using `T_1 = 300 K, p_1 = 90 atms, T_2 = 350 K, p_2 = 110 atm, V = 0.250 "litre"` `a = 1.87 "atm.litre"^2//"mole"^2, b = 0.045 litre//"mole"`. |
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