Question

One mole of an ideal gas undergoes isothermal irreversible expansion against constant
pressure 5 atm at 300k from 50m3 to 550m3 . Calculate work done into KJ /mole

Answer 1

$\huge\boxed{\underline{\mathcal{\red{❥A}\green{N}\pink{S}\orange{W}\blue{E}\purple{R}}}}$

Sea water is a mixture of salts and water which cannot be separated except by evaporation. Therefore, sea water is considered homogeneous.

Sea water also contains mud, decayed plant, etc., other than salts and water, so it is heterogeneous also.

Answer 2

Given:

One mole of an ideal gas undergoes isothermal irreversible expansion against constant pressure 5 atm at 300k from 50m3 to 550m3.

To find:

Work done in kJ/mole

Calculation:

The general expression for work done in reversible isothermal expansion :

$\boxed{ \rm \therefore \: W = - 2.303 \times nRT \: \log( \dfrac{ V_{2} }{V_{1}} )}$

Putting values in SI units :

$\rm \implies \: W = - 2.303 \times 1 \times 8.314 \times 300\: \times \log( \dfrac{550}{50} )$

$\rm \implies \: W = - 5744.14 \times \log( \dfrac{550}{50} )$

$\rm \implies \: W = - 5744.14 \times \log(11)$

$\rm \implies \: W = - 5744.14 \times 1.04$

$\rm \implies \: W = - 5981.9 \: J$

$\rm \implies \: W = - 5.98 \: kJ$

So , work done is -5.98 kJ/mol.