Ncert Exercises & Solutions

The pressure-volume work for an ideal gas can be calculated by using the expression $W = \int_{V_{i}}^{V_{f}} p_{ex} d V$ .

The work can also be calculated from the pV-plot by using the area under the curve within the specified limits.
An ideal gas is compressed (a) reversibly or (b) irreversibly from volume $V_{i}$ to $V_{f}$ .
The correct option is:

1. $W (reversible) = W (irreversible)$

2. $W (reversible) < W (irreversible)$

3. $W (reversible)> W (irreversible)$

4. $W (reversible) = W (irreversible) + p_{ex} . ∆ V$

Hint: For compression, work done for the reversible process is less than the irreversible process

The correct option is W (reversible) < W (irreversible). This is because the area under the curve is always more in irreversible compression as can be seen from the given figure.

A reaction at equilibrium is reversible. Conversely, a reaction that is not at equilibrium is irreversible. All real processes are irreversible.

Reversible processes produce the maximum amount of work

If a process does work on the surroundings, you get more work out of the process if it is done slowly. This is because less heat is lost to the surroundings. So a reversible process (infinitely slow) does the maximum work. There is one example of this that is easy to understand: the work done by a gas expanding inside a piston. We will see that the maximum amount of work is obtained when the gas is allowed to expand very slowly (reversibly).

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