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A piston-cylinder system contains 0.5 moles of an ideal gas initially at a pressure of 2 \mathrm{~atm} and a volume of 10 \mathrm{~L}.The gas undergoes an isothermal expansion at a constant temperature of 300 \mathrm{~K} until its volume doubles. Calculate the work done by the gas during the expansion and the amount of heat transferred.

Given:
          \text { Initial pressure }\left(P_{i}\right) =2 \mathrm{~atm}
          \text { Initial volume }\left(V_{i}\right) =10 \mathrm{~L}
          \text { Final volume }\left(V_{f}\right) =20 \mathrm{~L}
         \text { Number of moles }(n) =0.5 \mathrm{moles}
         \text { Gas constant }(R) =8.314 \mathrm{~J} /(\mathrm{mol} \mathrm{K})
          \text { Temperature }(T) =300 \mathrm{~K}

 

Option: 1

-20 L atm


Option: 2

-25  L atm


Option: 3

-24  L atm


Option: 4

-18  L atm


Answers (1)

best_answer

Step 1: Calculate the change in internal energy using the ideal gas law:

\mathrm{U=n C_{v} T}
Where:
         \mathrm{n }  is the number of moles
         \mathrm{C_{v} } is the molar heat capacity at constant volume
         \mathrm{T } is the change in temperature

  Since the process is isothermal,\mathrm{T=0 }.
\mathrm{U=0 \mathrm{~J} }

Step 2: Calculate the work done by the gas using the formula for expansion work:

\mathrm{W=-U=-P V }
Where:
\mathrm{p} is the pressure
V  is the change in volume
\mathrm{W=-\left(P_{i}\right)\left(V_{f}-V_{i}\right)}
Substitute the given values:

\mathrm{W=-(2 \operatorname{atm})(20 \mathrm{~L}-10 \mathrm{~L})}
\mathrm{W=-20 \mathrm{~L} \text { atm }}

Step 3: Calculate the amount of heat transferred using the first law of thermodynamics:
\begin{aligned} & U=Q-W \\ & Q=U+W \end{aligned}
Substitute the calculated value of U and W :
Q=0 \mathrm{~J}+(-20 \mathrm{~L} \text { atm })
Q=-20 \mathrm{~L} \text { atm }
Answer: During the isothermal expansion, the gas does -20 \mathrm{~L} atm of work, and -20 \mathrm{~L} atm of heat is transferred. Therefore, the correct option is  A.

Posted by

Kuldeep Maurya

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