22gm of carbondioxide are compressed isothermally and reversibaly 294k from pressure of 100kpa when work obtained is 1.4 KJ find the final pressure
Answers
Explanation:
Final pressure ?
Calculation:
General expression for reversible isothermal compression :
\boxed{\bf \: W = - 2.303 \times nRT \times log( \dfrac{P1}{P2} ) }W=−2.303×nRT×log(P2P1)
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} )⟹1300=−2.303×(4422)×8.314×298×log(P2P1)
\sf\: \implies 1300 = - 2.303 \times \dfrac{1}{2} \times 8.314 \times 298 \times log( \dfrac{P1}{P2} )⟹1300=−2.303×21×8.314×298×log(P2P1)
\sf\: \implies 1300 = - 2852.9 \times log( \dfrac{P1}{P2} )⟹1300=−2852.9×log(P2P1)
\sf\: \implies log( \dfrac{P1}{P2} ) = - 0.45⟹log(P2P1)=−0.45
\sf\: \implies \dfrac{P1}{P2} = {10}^{ - 0.45}⟹P2P1=10−0.45
\sf\: \implies \dfrac{P1}{P2} = 0.354⟹P2P1=0.354
\sf\: \implies \dfrac{100 \times {10}^{3} }{P2} = 0.354=10−0.45
\sf\: \implies \dfrac{P1}{P2} = 0.354⟹P2P1=0.354
\sf\: \implies \dfrac{100 \times {10}^{3} }{P2} = 0.354⟹P2100×103=0.354
\sf\: \implies \dfrac{{10}^{5} }{P2} = 0.354⟹P2105=0.354
\sf\: \implies P2 = \dfrac{ {10}^{5}}{0.354}⟹P2=0.354105
\sf\: \implies P2 = \dfrac{ {10}^{2}}{0.354}\:kPa⟹P2=0.354102kPa
\sf\: \implies P2 = 282.4 \: kPa⟹P2=282.4kPa
So, final pressure is 282.4 kPa.