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  • A gas confined in a piston-cylinder system undergoes a reversible expansion against a constant external pressure of 1 atm. The initial volume of the gas is <img alt="10 \mathrm{~L}" src="https://en

A gas confined in a piston-cylinder system undergoes a reversible expansion against a constant external pressure of 1 atm. The initial volume of the gas is 10 \mathrm{~L},and the final volume is 30 \mathrm{~L}.During the expansion, 1500 \mathrm{~J} of heat is added to the gas. Calculate the work done by the gas during the expansion and the change in internal energy of the gas.

Given:
          \text { External pressure }\left(P_{\text {ext }}\right) =1\mathrm{~atm}
          \text { Initial volume }\left(V_{i}\right) =10 \mathrm{~L}
          \text { Final volume }\left(V_{f}\right) =30 \mathrm{~L}
          \text { Heat added }(Q) =1500 \mathrm{~J}
 
            

Option: 1

2450 \mathrm{~J}


Option: 2

2300 \mathrm{~J}


Option: 3

2800 \mathrm{~J}


Option: 4

2500 \mathrm{~J}


Answers (1)

best_answer

Step 1: Calculate the work done by the gas during the reversible expansion against a constant external pressure:

W=-P_{\text {ext }} \Delta V
Where:
\begin{aligned} & P_{\text {ext }} \text { is the constant external pressure } \\ & \Delta V=V_{f}-V_{i} \text { is the change in volume } \end{aligned}

Substitute the given values:

W=-(1 \mathrm{~atm}) \times(30 \mathrm{~L}-10 \mathrm{~L})
W=-20 \mathrm{~L} \mathrm{~atm}

Step 2: Calculate the change in internal energy using the first law of thermodynamics:
\Delta U=Q-W
Where:

\begin{aligned} & Q \text { is the heat added } \\ & W \text { is the work done } \end{aligned}

Substitute the given values:

\Delta U=1500 \mathrm{~J}-(-20 \mathrm{~L} \mathrm{~atm}) \times(101.325 \mathrm{~J} / \mathrm{L} \mathrm{atm})
\Delta U \approx 2500 \mathrm{~J}

Answer: During the reversible expansion, the gas does -20 \mathrm{~L}atm of work, and the change in internal energy of the gas is approximately 2500 \mathrm{~J}.

Posted by

Gautam harsolia

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