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Assume an ideal gas obeys $\mathrm{P V= constant}$ . This gas confined in a piston-fitted cylinder has an initial volume and pressure of 2 liters and $\mathrm{1 \: atm}$ , respectively, and is allowed to expand to finally occupy 6 liters. Calculate the work done by the system.

Option: 1
$105 \mathrm{~J}$

Option: 2
$206 \mathrm{~J}$

Option: 3
$110 \mathrm{~J}$

Option: 4
$130 \mathrm{~J}$

Answers (1)

best_answer

Step 1: Express $\mathrm{P}$ in terms of $\mathrm{V}$ using the equation $\mathrm{PV=constant}$

$\mathrm{P_1 V_1 =P_2 V_2 }$

$\mathrm{P_2 =\frac{P_1 V_1}{V_2}}$

Step 2: Calculate the work done using the integral.

$\mathrm{ W=\int_{V_1}^{V_2} \frac{P_1 V_1}{V} d V }$

Step 3: Evaluate the definite integral.

$\mathrm{ W=P_1 V_1[\ln |V|]_{V_1}^{V_2} }$

Step 4: Substitute the given values and calculate the work.

$\mathrm{ W=(1 \mathrm{~atm}) \cdot(2 \text { liters }) \cdot[\ln \mid 6 \text { liters }|-\ln | 2 \text { liters } \mid] }$

Calculating the numerical value of the work:

$\mathrm{ W \approx(1 \mathrm{~atm}) \cdot(2 \text { liters }) \cdot(1.7918-0.6931) \approx 1.0986 \mathrm{~L} \text {-atm } \approx 105.3 \mathrm{~J} }$

So, the work done by the system during this expansion is approximately $\mathrm{ 105.3\: joules. }$

So, the correct option is 1.

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