Question

If one mole of an ideal gas expands isothermally at 37 degrees from 15 litres to 25 litres the maximum work obtained is

Answer 1

Answer:

$W_{rev} =-2.303nRTlog\frac{V_{2} }{V_{1} }$

                    =$-2.303*1*0.0821*(273+37)log\frac{25}{15}$

                    =$-2.303*0.0821*310log\frac{5}{3}$

                    =13J

                    =12.87J

Answer 2

The maximum work obtained is -1892 J

Explanation:

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

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

where,

W = work done

n = number of moles = 1 mole

R = Gas constant = 8.314 J/mol.K

T = Temperature of the gas = $37^oC=[37+273]K=310K$

$V_1$ = initial volume = 15 L

$V_2$ = final volume = 25 L

Putting values in above equation, we get:

$W=-2.303\times 1mol\times 8.314J/mol.K\times 310K\times \log(\frac{25}{15})\\\\W=-1892J$

Maximum work obtained is -1892 J

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