Show that, for a classical ideal gas, S/NK = ln(Q_1/N + T(partial differential l

Question

Show that, for a classical ideal gas, S/NK = ln(Q_1/N + T(partial differential ln Q_1/partial differential T_P. If an ideal monatomic gas is expanded adiabatically to twice its initial volume, what will the ratio of the final pressure to the initial pressure be? If during the process some heat is added to the system, will the final pressure be higher or lower than in the preceding case? Support your answer by deriving the relevant formula for the ratio P_f/P_i. The volume of a sample of helium gas is increased by withdrawing the piston of the containing cylinder. The final pressure P_f is found to be equal to the initial pressure P_i times (V_i/V_f)^1.2, V_i and V_f being the initial and final volumes. Assuming that the product PV is always equal to 2/3 U, will (i) the energy and (ii) the entropy of the gas increase, remain constant, or decrease during the process? If the process were reversible, how much work would be done and how much heat would be added in doubling the volume of the gas? Take P_i = 1 atm and V_i = 1 m^3. Determine the work done on a gas and the amount of heat absorbed by it during a compression from volume V_1 to volume V_2, following the law PV^n = const. If the "free volume" V^bar of a classical system is defined by the equation. V^bar N = integral e^{U^bar - U(q_1}/kT Pi_i=1^N d^3 q_i, where U^bar is the average potential energy of the system and U(q_i the actual potential energy as a function of the molecular configuration, then show that

Explanation / Answer

3.9 First of all for an adiabatic process we know that PV^ =K =constant

In this case Vf=2Vi

then for ideal monoatomic gas

PiVi^=PfVf^

or PiVi^= Pf[2Vi]^

or Pf/Pi =Vi^/ 2^Vi^ =1/2^

According to the basic law heat cannot be added/released into adibatic process

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