# for a process to be isothermal temperature must be constant....means there must be sufficient time for the heat to be exchanged between system and surroundings so that temp remains constant.... means the process needs to be slow if it is to be isothermal and slow processes are always reversible....so there must be nothing like 'irreversible isothermal' cause it wont be possible.

### Asked by Nishtha Gupta | 11th Jan, 2014, 11:56: AM

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Isothermal Reversible Processes are possible.

A reversible process is one that can be halted at any stage and reversed. In a reversible process the system is at equilibrium at every stage of the process. An irreversible process is one where these conditions are not fulfilled.

If p_{int} > p_{ext} in an expansion process then the process is irreversible because the system does not remain at equilibrium at every stage of the process. (There will be turbulence and temperature gradients).

Example:

In the experimental process, the gas is compressed isothermally from an initial pressure of 1.0 atmosphere and volume of 30.0 mL to a final pressure of 3.0 atmospheres and a volume of 10.0 mL. The temperature throughout is 20.0C. The isothermal compression of the gas was performed in two ways, one irreversible and one reversible.

Irreversible isothermal compression:

The irreversible compression of the gas proceeds in one step against a constant external pressure of 3.0atm. The work is given by

w = -P_{ext}ΔV = 0.060 Latm

The work done on the gas is positive because the gas is compressed; the surroundings do positive work on the gas.

How do we compute the heat added to the gas? Indirectly! We note that the process is isothermal. The total energy of the ideal gas is

ΔE = n c_{v}ΔT

and since the temperature is constant, ΔT=0 and ΔE=0. Using the first law of thermodynamics

ΔE = q + w

We have q=-w=-0.060 Latm.

The heat added to the system is negative because the gas is compressed. Why? In the absence of heat transfer, the gas will gain internal energy in the compression and its temperature will increase. To keep the temperature of the gas constant, the gas must transfer a positive amount of heat to the surroundings. So the heat added to the compressed gas is negative, keeping the temperature constant.

Finally, for an ideal gas

ΔH = n c_{p}ΔT

and since the temperature is constant, ΔT=0 and it follows that ΔH=0.

Reversible isothermal compression:

The reversible compression of the gas proceeds in an infinite number of infinitesimal steps, where the external pressure applied is always essentially equal to the internal pressure of the gas (or larger only by an infinitesimal amount). The work is given by

w = -nRT ln(V_{2}/V_{1}) = -(0.030 Latm) ln(0.010L/0.030L) = 0.033 Latm.

The work done on the gas is positive because the gas is compressed; the surroundings do positive work on the gas. Note that the work done on the gas is greater for the irreversible process

w_{irr} > w_{rev}

which is true in general.

We compute the heat added to the gas as we did for the irreversible process. Indirectly! Again the change in energy of the system for the isothermal process is

ΔE = 0

and we have q=-w=-0.033 Latm.

Note that the change in energy must be the same for the reversible and irreversible processes because the initial and final states of the gas are the same. As the change in energy depends only on the initial and final states of the system, it is independent of the path followed. It follows that ΔH=0.

Isothermal Reversible Processes are possible.

A reversible process is one that can be halted at any stage and reversed. In a reversible process the system is at equilibrium at every stage of the process. An irreversible process is one where these conditions are not fulfilled.

If p_{int} > p_{ext} in an expansion process then the process is irreversible because the system does not remain at equilibrium at every stage of the process. (There will be turbulence and temperature gradients).

Example:

In the experimental process, the gas is compressed isothermally from an initial pressure of 1.0 atmosphere and volume of 30.0 mL to a final pressure of 3.0 atmospheres and a volume of 10.0 mL. The temperature throughout is 20.0C. The isothermal compression of the gas was performed in two ways, one irreversible and one reversible.

Irreversible isothermal compression:

The irreversible compression of the gas proceeds in one step against a constant external pressure of 3.0atm. The work is given by

w = -P_{ext}ΔV = 0.060 Latm

The work done on the gas is positive because the gas is compressed; the surroundings do positive work on the gas.

How do we compute the heat added to the gas? Indirectly! We note that the process is isothermal. The total energy of the ideal gas is

ΔE = n c_{v}ΔT

and since the temperature is constant, ΔT=0 and ΔE=0. Using the first law of thermodynamics

ΔE = q + w

We have q=-w=-0.060 Latm.

The heat added to the system is negative because the gas is compressed. Why? In the absence of heat transfer, the gas will gain internal energy in the compression and its temperature will increase. To keep the temperature of the gas constant, the gas must transfer a positive amount of heat to the surroundings. So the heat added to the compressed gas is negative, keeping the temperature constant.

Finally, for an ideal gas

ΔH = n c_{p}ΔT

and since the temperature is constant, ΔT=0 and it follows that ΔH=0.

Reversible isothermal compression:

The reversible compression of the gas proceeds in an infinite number of infinitesimal steps, where the external pressure applied is always essentially equal to the internal pressure of the gas (or larger only by an infinitesimal amount). The work is given by

w = -nRT ln(V_{2}/V_{1}) = -(0.030 Latm) ln(0.010L/0.030L) = 0.033 Latm.

The work done on the gas is positive because the gas is compressed; the surroundings do positive work on the gas. Note that the work done on the gas is greater for the irreversible process

w_{irr} > w_{rev}

which is true in general.

We compute the heat added to the gas as we did for the irreversible process. Indirectly! Again the change in energy of the system for the isothermal process is

ΔE = 0

and we have q=-w=-0.033 Latm.

Note that the change in energy must be the same for the reversible and irreversible processes because the initial and final states of the gas are the same. As the change in energy depends only on the initial and final states of the system, it is independent of the path followed. It follows that ΔH=0.

### Answered by Vaibhav Chavan | 15th Jan, 2014, 11:24: AM

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