Question

Calculate the amount of work done by 2 mole of an ideal gas at 298 K in reversible expansion from 10 litre to 20 litre.

Answer 1

Hint: The ideal gas law equation can be defined as the equation of state of a hypothetical ideal gas. It gives us a good approximation about the behavior of many gases under many conditions but it goes through some limitations too.

Complete step by step answer:
An isothermal processes are those processes in which the temperature of the system remains constant throughout the reaction and reversibly expansion of the gas takes place in a finite number of infinitesimally small intermediate steps. Amount of work done in reversibly expansion is given by the formula:
$W = -2.303RT\log \dfrac{{{V}_{2}}}{{{V}_{1}}}$
Here n = 2 (given), R = 8.314 (Fixed value of gas constant), T = 298 K, ${{V}_{2}}$ = 20L and ${{V}_{1}}$ = 10L (given)
Put the values in the work done formula
$W = -2.303\times 8.314\times 298\log \frac{20}{10}$
$W = -2.303\times 8.314\times 298\times 0.310$
$W = -3434.9J$
- Hence the amount of work done by 2 mole of an ideal gas at 298 K in reversible expansion from 10 litre to 20 litre is $-3434.9J$.
- Negative sign indicates that work is done by the system whereas positive sign of work done indicates work is done on the system.

Note: The state of an ideal gas is determined by the macroscopic and microscopic parameters known as pressure, volume, temperature. But in reality ideal gas does not exist; it is just a hypothetical gas proposed to simplify the calculations.