Question

1)

Three parts

A) True or false. Briefly explain why.

Given the statement: 1 mole of vapor mixture consisting of A and B is enclosed in a thermally conductive cylinder by a frictionless piston. The mixture is ignited to start a chemical reaction between A and B. This causes the piston to move up and down violently. After the gas returns to its initial pressure P₀ and temperature T₀:

i) TRUE OR FALSE: The heat gas releases to the environment is equal to the decrease of its entropy times T₀.

ii) TRUE OR FALSE: The heat gas releases to the environment is equal to the decrease of its enthalpy.

B) 1 mole of ideal gas is enclosed in a cylinder by a piston. Both cylinder and piston are thermally insulated. What are the signs of the following quantities of the gas (>0, =0, or <0):

i) Piston is moved very slowly to double the gas volume

ΔT

ΔU

ΔS

ΔP

ii) Piston is moved very rapidly to double the gas volume (Hint: if piston moves much faster than gas expansion, it temporarily creates vacuum insiade the cylinder) What are the signs of the following quantities of the gas (>0, =0, or <0)

ΔT

ΔU

ΔS

ΔP

C) Use the relation between entropy and the number of microstates to briefly explain why solids usually absorb heat upon melting and liquids absorb heat upon boiling at the equilibrium transition temperatures,

Answer 1

1.A.i) True because for a reversible process heat transfer is equal to change in entropy times temperature T.

ii) False because initial and final state are same so change of state function will be zero.

B)I)∆T= 0 since process will be isothermal

∆U=O since for ideal gas U is a function of temperature only.

∆S<O because work will be done by the system.so ∆q will be negative.

∆P<0 since for isothermal process pressure decreases for increasing volume.

ii)∆T>0 since the process will be adiabatic.

∆U>0 since the process will be adiabatic.

∆S=0 since dq will be zero

∆P>0 since the process will be adiabatic.