In the V-T graph shown, an ideal gas undergoes a cyclic process ABCDA. AB curve is a rectangular hyperbola and T1 =
300 K and T2 = 500 K. If the ratio of maximum to minimum volume is 20:9, then the ratio of maximum to minimum
pressures in the cycle is
a)5/3
b)5/4
c)25/9
d)20/9
Answers
(A) is the correct option as:-
From the figure of option (A), V B =V C and T A =T C
A-C is isothermal process
B-C is constant volume process on PV diagram in solutions
A-C is a rectangular hyperbola
B-C is constant volume process
B is the maximum temperature point.
Answer:
None of these.
Explanation:
In order to solve this problem, we need to use the gas laws and the ideal gas equation to relate the volume and pressure of the gas to its temperature. The ideal gas equation is given by
PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature.
From the graph, we can see that the AB curve is a rectangular hyperbola. This means that PV = constant for this curve. Therefore, we can write:
P1V1 = P2V2
where P1 and V1 are the pressure and volume at point A, and P2 and V2 are the pressure and volume at point B.
We also know that the temperature at point A is 300 K, and the temperature at point B is 500 K. Therefore, we can write:
V1/V2 = T2/T1 = 5/3
We are given that the ratio of maximum to minimum volume is 20:9. Therefore:
Vmax/Vmin = 20/9
Let Vmax be the volume at point A and Vmin be the volume at point C. Then:
Vmax/Vmin = Vmax/V2 * V2/Vmin = Vmax/V2 * (Vmax/V2)/(Vmax/Vmin) = (Vmax/V2)^2
Therefore:
(Vmax/V2)^2 = 20/9
Vmax/V2 = sqrt(20/9) = 2sqrt(5)/3
Similarly, from the ideal gas equation, we know that:
PmaxVmax/Tmax = PminVmin/Tmin
Since Tmax = T1 and Tmin = T2, we have:
PmaxVmax = PminVmin(T1/T2)
Let Pmax be the pressure at point A and Pmin be the pressure at point C. Then:
Pmax/Pmin = Vmin/Vmax * T1/T2 = (9/20) * (300/500) = 27/100
Therefore, the ratio of maximum to minimum pressures in the cycle is 27/100, which is not one of the given answer options.
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