How does pressure affect enthalpy?
Answer
614.7k+ views
Hint: Enthalpy is the heat content of a system as a function of entropy and pressure.
Enthalpy is a central factor in thermodynamics. It is the total heat contained in the system.
Complete step by step answer:
Enthalpy (H) is defined as
$H = U{\text{ }} + {\text{ }}PV $
$ = q + w + P\Delta V + V\Delta P + \Delta P\Delta V $
$ = T\Delta S - P\Delta V + P\Delta V + V\Delta P + \Delta P\Delta V$
$H = T\Delta S + V\Delta P + \Delta P\Delta V $
U is internal energy and P and V are pressure and volume respectively. At constant volume, the heat emitted or absorbed during a reaction equals the internal energy of a system. At constant pressure; heat emitted or absorbed during a chemical reaction equals the enthalpy of the system. Both these points again go back to the original equation.
As the pressure increases $$(\Delta P > 0)$$, so does enthalpy, and vice versa. Thus, more compressed molecules such as solids have greater intermolecular forces than less compressed molecules such as liquids or gases; their interactions are harder to separate.
Additional information:
An ideal gas is a theoretical gas.
The ideal gas law is the equation of state for an ideal gas, given by:
$$PV = nRT$$
Where
P is the pressure
V is the volume
N is the amount of substance of the gas (in moles)
R is the gas constant $$\left( {0.08206\;L\cdot atm\cdot{K^{ - 1}}\cdot mo{l^{ - 1}}} \right)$$
T is the absolute temperature.
Thus, enthalpy does not depend on pressure at constant T and it is a function of temperature only.
Note: Preparing for a subject requires you to be clear on every topic included in it. You should be able to relate enthalpy definition and derivation with the subject matter at hand. Being familiar with the concepts and numerical in that specific chapter is the key to complete the exam.
Enthalpy is a central factor in thermodynamics. It is the total heat contained in the system.
Complete step by step answer:
Enthalpy (H) is defined as
$H = U{\text{ }} + {\text{ }}PV $
$ = q + w + P\Delta V + V\Delta P + \Delta P\Delta V $
$ = T\Delta S - P\Delta V + P\Delta V + V\Delta P + \Delta P\Delta V$
$H = T\Delta S + V\Delta P + \Delta P\Delta V $
U is internal energy and P and V are pressure and volume respectively. At constant volume, the heat emitted or absorbed during a reaction equals the internal energy of a system. At constant pressure; heat emitted or absorbed during a chemical reaction equals the enthalpy of the system. Both these points again go back to the original equation.
As the pressure increases $$(\Delta P > 0)$$, so does enthalpy, and vice versa. Thus, more compressed molecules such as solids have greater intermolecular forces than less compressed molecules such as liquids or gases; their interactions are harder to separate.
Additional information:
An ideal gas is a theoretical gas.
The ideal gas law is the equation of state for an ideal gas, given by:
$$PV = nRT$$
Where
P is the pressure
V is the volume
N is the amount of substance of the gas (in moles)
R is the gas constant $$\left( {0.08206\;L\cdot atm\cdot{K^{ - 1}}\cdot mo{l^{ - 1}}} \right)$$
T is the absolute temperature.
Thus, enthalpy does not depend on pressure at constant T and it is a function of temperature only.
Note: Preparing for a subject requires you to be clear on every topic included in it. You should be able to relate enthalpy definition and derivation with the subject matter at hand. Being familiar with the concepts and numerical in that specific chapter is the key to complete the exam.
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