Question

1. (a) In a different galaxy far from ours, let us assume that the "ideal" gas laws are as follows: (i) At constant temperature, pressure is inversely proportional to the square volume. (ii) At constant pressure, the volume varies directly with the temperature to power. (ii) At 273.16 K and 1 atm pressure, one mole of an "ideal" gas is found to occupy 22.414 liters. Under these conditions, show that pT constant, and obtain the form of the "ideal" gas equation-of-state in this galaxy. (b) The coefficient of thermal expansion G is defined to be G-(1/V)[dVdT]p. Obtain G for the above "ideal gas as a function of T alone. of the the 2/3

Answer 1

Solution (A) Expressing the three laws in mathematical form, we have k(T) where k(T) is an unknown function of T. where C(P) is an unknown function of P. Combining Eqs. 1 and 2 gives k(T) CYP)7% Separating the variables, we obtain k(T) which must hold for all values of P and T. Consequently, we must have k(T) T43 PC2(p)--= G-a constant. This gives CYP) = Substituting Eq. 6 into Eq. 2 yields G12T23 Rearranging Eq. 7 produces T23 Raising both sides of Eq. 8 to the sixth power, we obtairn - G3a constant. T- Substituting the third law gives (1 atm) (22.414L mol-n mol) (273.15 K) (10) The “ideal-gas law" is, therefore, where R-0.022778 atm L6 mol 6K-4

(B) To obtain the coefficient of thermal expansion, we need (aV/aT). Solving Eq. 11 for V, we obtain (12) Differentiating both sides with respect to T at constant P gives 6Vs (13) aT Thus, av2n RT3 (14) α is, therefore, given by (15) oT 3P3V Replacing Py6 in the denominator with n RT from Eq. 11, we obtain (16) which is the required expression. We may also solve this problem by rewriting the expression for α: (17) aT Taking logarithms of both sides of Eq. 11 gives 31n P + 61n V=61n n+In R + 4 In T (18) or [6 In n In R + 4 In T-31n P1 (19) from which we immediately obtain a In V (20) 6T 3T obtained in Eq. 16. which is identical to the expression