Question

work done when 2 moles of an ideal gas is compressed from a volume of 5 m3 to 1 dm3 at 300K under a pressure of 100 Pa

Answer 1

Answer: The amount of work done by the ideal gas is 42494.8 J

Explanation:

To calculate the work done for isothermal, reversible compression process, we use the equation:

$W=-2.303nRT\log(\frac{V_2}{V_1})$

where,

W = work done

n = number of moles = 2 mole

R = Gas constant = 8.314 J/mol.K

T = Temperature of the gas = 300 K

$V_1$ = initial volume = $5m^3=5000dm^3$    (Conversion factor:  $1m^3=1000dm^3$ )

$V_2$ = final volume = $1dm^3$

Putting values in above equation, we get:

$W=-2.303\times 2mol\times 8.314J/mol.K\times 300K\times \log(\frac{1}{5000})\\\\W=42494.8J$

Hence, the amount of work done by the ideal gas is 42494.8 J