(e) The system starts at temperature To, magnetic field Ho, and pressure po. The magnetic field is changed quasistatically, adiabatically, and isothermally to a new value H. (This requires adjusting the pressure and hence the volume - to keep T constant without adding heat.) Find the new pressure p in terms of H, Ho, To, Po, and constants. (f) The system starts at temperature To and magnetic field Ho. The magnetic field is changed adiabatically, quasistatically, and at constant pressure, to a new value H. Inte- grate the appropriate differential equation to find an equation relating the new temper- ature T to H, Ho, To, and constants. (This equation can be solved for T as a function of H, Ho, To, but you do not need to do it.) [Hint: Define a new variable f = aH²/(2T²), find df, and use it to simplify your differential equation.]
(e) The system starts at temperature To, magnetic field Ho, and pressure po. The magnetic field is changed quasistatically, adiabatically, and isothermally to a new value H. (This requires adjusting the pressure and hence the volume - to keep T constant without adding heat.) Find the new pressure p in terms of H, Ho, To, Po, and constants. (f) The system starts at temperature To and magnetic field Ho. The magnetic field is changed adiabatically, quasistatically, and at constant pressure, to a new value H. Inte- grate the appropriate differential equation to find an equation relating the new temper- ature T to H, Ho, To, and constants. (This equation can be solved for T as a function of H, Ho, To, but you do not need to do it.) [Hint: Define a new variable f = aH²/(2T²), find df, and use it to simplify your differential equation.]
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Unsure on how to do the following problem involving statistical
![(e) The system starts at temperature To, magnetic field Ho, and pressure po. The magnetic
field is changed quasistatically, adiabatically, and isothermally to a new value H. (This
requires adjusting the pressure – and hence the volume – to keep T constant without
adding heat.) Find the new pressure p in terms of H, Ho, To, Po, and constants.
(f) The system starts at temperature To and magnetic field Ho. The magnetic field is
changed adiabatically, quasistatically, and at constant pressure, to a new value H. Inte-
grate the appropriate differential equation to find an equation relating the new temper-
ature T to H, H0, To, and constants. (This equation can be solved for T as a function of
H, Ho, To, but you do not need to do it.) [Hint: Define a new variable f = aH² /(2T²),
find df, and use it to simplify your differential equation.]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F591ac009-156d-459c-82fc-cac6b9112e8c%2Fcd53027e-770f-4d92-bb64-84475f043de3%2Ffqch0wb_processed.png&w=3840&q=75)
Transcribed Image Text:(e) The system starts at temperature To, magnetic field Ho, and pressure po. The magnetic
field is changed quasistatically, adiabatically, and isothermally to a new value H. (This
requires adjusting the pressure – and hence the volume – to keep T constant without
adding heat.) Find the new pressure p in terms of H, Ho, To, Po, and constants.
(f) The system starts at temperature To and magnetic field Ho. The magnetic field is
changed adiabatically, quasistatically, and at constant pressure, to a new value H. Inte-
grate the appropriate differential equation to find an equation relating the new temper-
ature T to H, H0, To, and constants. (This equation can be solved for T as a function of
H, Ho, To, but you do not need to do it.) [Hint: Define a new variable f = aH² /(2T²),
find df, and use it to simplify your differential equation.]
![A paramagnetic gas is a system with three independent variables, which may be chosen to be
any three of S, p, T, V, M, and H. (Here M is magnetization of the gas, and H is the external
magnetic field, not the enthalpy.) The basic equation is
dE = TdS – pdV + HdM
where the last term represents the increase in energy of the molecules when their magnetic
dipole moments are increased in the presence of an external magnetic field. [Aside: To get the
sign of the HdM term right, one has to consider time-dependent effects (i.e., electromagnetic
induction) due to changing magnetic moments. One would start with Jackson, Eqn. (6.15),
H + 4M, where M is the magnetization (magnetic dipole moment per unit
and use B =
volume), and H is the external field whose sources are fixed.]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F591ac009-156d-459c-82fc-cac6b9112e8c%2Fcd53027e-770f-4d92-bb64-84475f043de3%2F896js5r_processed.png&w=3840&q=75)
Transcribed Image Text:A paramagnetic gas is a system with three independent variables, which may be chosen to be
any three of S, p, T, V, M, and H. (Here M is magnetization of the gas, and H is the external
magnetic field, not the enthalpy.) The basic equation is
dE = TdS – pdV + HdM
where the last term represents the increase in energy of the molecules when their magnetic
dipole moments are increased in the presence of an external magnetic field. [Aside: To get the
sign of the HdM term right, one has to consider time-dependent effects (i.e., electromagnetic
induction) due to changing magnetic moments. One would start with Jackson, Eqn. (6.15),
H + 4M, where M is the magnetization (magnetic dipole moment per unit
and use B =
volume), and H is the external field whose sources are fixed.]
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