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  • A gas in a piston-cylinder system is compressed from an initial volume of to a final volu

A gas in a piston-cylinder system is compressed from an initial volume of 20 \mathrm{~L} to a final volume of 5 \mathrm{~L}.During the compression, the pressure of the gas increases from 1 \mathrm{~atm} to 4 \mathrm{~atm}. Calculate the work done during the compression and the change in enthalpy of the gas.

Given:
       \begin{aligned} \text { Initial volume }\left(V_{i}\right) & =20 \mathrm{~L} \\ \text { Final volume }\left(V_{f}\right) & =5 \mathrm{~L} \\ \text { Initial pressure }\left(P_{i}\right) & =1 \mathrm{~atm} \\ \text { Final pressure }\left(P_{f}\right) & =4 \mathrm{~atm} \\ \text { Gas constant }(R) & =0.0821 \mathrm{~L} \mathrm{~atm} /(\mathrm{mol} \mathrm{K}) \end{aligned}

Option: 1

37.5 \mathrm{~L} atm


Option: 2

25.8 \mathrm{~L} atm


Option: 3

29.5 \mathrm{~L}atm


Option: 4

34.2 \mathrm{~L} atm


Answers (1)

best_answer

Step 1: Calculate the work done during the compression using the formula for pressure-volume work:
W=-P \Delta V

Where:
P is the average pressure during the compression
\Delta V=V_{f}-V_{i}  is the change in volume

Calculate the average pressure:
\mathrm{P=\frac{P_{i}+P_{f}}{2}=2.5 \mathrm{~atm}}
Calculate \mathrm{\Delta V:}
\mathrm{\Delta V=V_{f}-V_{i}=5 \mathrm{~L}-20 \mathrm{~L}=-15 \mathrm{~L}}

Calculate the work done:
\mathrm{W=-(2.5 \mathrm{~atm}) \times(-15 \mathrm{~L})=37.5 \mathrm{~L} \mathrm{~atm}}

Step 2: Calculate the change in enthalpy \mathrm{(\Delta H)} using the first law of thermodynamics:
\mathrm{\Delta H=Q+W}

Where:

\mathrm{ Q \text { is the heat added or released } }
\mathrm{ W \text { is the work done } }

Since the problem doesn't provide information about heat transfer \mathrm{ (Q) }, we will assume that the process is adiabatic \mathrm{ (Q=0) }.
Substitute the calculated value of \mathrm{ W: }
\mathrm{ \Delta H=0+37.5 \mathrm{~L} \mathrm{~atm}=37.5 \mathrm{~L} \mathrm{~atm} }

Answer: The work done during the compression is \mathrm{ 37.5 \mathrm{~L} } atm, and the change in enthalpy \mathrm{ (\Delta H) }  is also \mathrm{ 37.5 \mathrm{~L} } atm.

Posted by

shivangi.bhatnagar

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