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At 300\: \mathrm{K\: cm} ideal gas expands internally from It to \mathrm{2 t} against a constant pressure of \mathrm{1.5 \mathrm{~atm}}. What is the change in enthalpy of Surroundings if the gas was \mathrm{1 mol\: in \, \mathrm{JK}^{-1}} \mathrm{(1L\: atm =101.3 \mathrm{~J} )} and the expansion is isothermal.
 

Option: 1

-0.5 \mathrm{JK}^{-1}
 


Option: 2

\mathrm{-2\: J K^{-1}}
 


Option: 3

\mathrm{+1 J K^{-1}}
 


Option: 4

\mathrm{+0.5 \mathrm{JK}^{-1}}


Answers (1)

best_answer

For isothermal \mathrm{\triangle H=O}

\mathrm{9 =-\omega=P_e \times \Delta H }

    \mathrm{=1.5(2-1) }

   \mathrm{=1.5\: \mathrm{L\: atm} }

\mathrm{=-1.5 \times 101.3 \mathrm{~J}=\text { Heat released } }

\mathrm{ \Delta S =\frac{-1.5 \times 101.3}{300}=-0.5065 \mathrm{JK}^{-1}}

Hence option 1 is correct.






 

Posted by

seema garhwal

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