Question

22 g of CO2 are compressed isothermally and reversibly at 298 K from initial pressure of 100 kPa when the work obtained is1.3 kJ . Find the final pressure.​


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Answer 1

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Answer 2

Given:

22 g of CO2 are compressed isothermally and reversibly at 298 K from initial pressure of 100 kPa when the work obtained is1.3 kJ .

To find:

Final pressure ?

Calculation:

General expression for reversible isothermal compression :

$\boxed{\bf \: W = - 2.303 \times nRT \times log( \dfrac{P1}{P2} ) }$

Putting the available values in SI units:

$\sf\: \implies 1300 = - 2.303 \times ( \dfrac{22}{44} ) \times 8.314 \times 298 \times log( \dfrac{P1}{P2} )$

$\sf\: \implies 1300 = - 2.303 \times \dfrac{1}{2} \times 8.314 \times 298 \times log( \dfrac{P1}{P2} )$

$\sf\: \implies 1300 = - 2852.9 \times log( \dfrac{P1}{P2} )$

$\sf\: \implies log( \dfrac{P1}{P2} ) = - 0.45$

$\sf\: \implies \dfrac{P1}{P2} = {10}^{ - 0.45}$

$\sf\: \implies \dfrac{P1}{P2} = 0.354$

$\sf\: \implies \dfrac{100 \times {10}^{3} }{P2} = 0.354$

$\sf\: \implies \dfrac{{10}^{5} }{P2} = 0.354$

$\sf\: \implies P2 = \dfrac{ {10}^{5}}{0.354}$

$\sf\: \implies P2 = \dfrac{ {10}^{2}}{0.354}\:kPa$

$\sf\: \implies P2 = 282.4 \: kPa$

So, final pressure is 282.4 kPa.