2) The following PV diagram for 1 mole of an ideal gas contains three segments:

Question

2) The following PV diagram for 1 mole of an ideal gas contains three segments: A to B is at constant pressure, B to C is at constant volume, and C to A is at constant temperature. The pressure la 3 x 106 Pa and the volume VA = point B is VB 10.314 m3. Furthermore, the individual mass of each of these particular ideal gas particles is 1.381 × 10-23 kg. Oh, by the way, 1.381 × 10-23 is also the value of the Boltzmann constant k in J/K. Also, 8.314 just happens to be the value of the gas constant in J/(mole K) a) Determine the work done by the gas in the segment AB. b) Recall, Kinetic theory (or the Equipartition Theorem) relates the root-mean-square speed vrmsVfor an ideal gas molecule to the temperature. Determine the root- mean-square speed for a molecule of this gas at point C 8.314 m3. In addition, the volume at nlmol m 1381 x1023 k A Constan Ressure B VR-10, 314 m 3 PA -23 R 331 mol K a. Constant Prssue A2 3 8 C Ve=10,314 (3X10") (a)

Explanation / Answer

a)Work done by the gas in an isobaric process = P*dV

=3*10^6 * (10.314 - 8.314)

=6*10^6 J

b)For the second part of the question,

Vrms = (3RT/M)^0.5

where M is the molar mass. Since the process BC is isochoric, by ideal gas law,

Pb/Tb = Pc/Tc

let the temperature at B be x.so, using the ideal gas law,

PV = nRT

or (3*10^6/(1.01*10^5)) * 10.314*10^3 = 1*(1/12)*T

or T = 3.67*10^6 K

Let the molar mass of the gas be M.

so, M= 6.023*10^23*1.381*10^-23

or M=8.31 g/mol

let the temperature at A be T.so using the ideal gas law,

PV = nRT

or (3*10^6/(1.01*10^5)) * 8.314*10^3 = 1*(1/12)*T

or Ta = Tc = 2.96*10^6 K

so Vrms is given by (3RT/M)^0.5

=(3*8.314*2.96*10^6/(8.31*10^-2))^0.5

=29806.5 m/s

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