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A gas confined within a piston-cylinder assembly undergoes a process where it expands at constant pressure of $2 \mathrm{~atm}$ from a volume of $\mathrm{0.02 \mathrm{~m}^3\: to\: 0.04 \mathrm{~m}^3}$ During this process, $1500 \mathrm{~J}$ of heat is added to the gas. Calculate the work done by the gas and its change in internal energy.

Option: 1
$20265,-18765 \mathrm{~J}$

Option: 2
$15469 \mathrm{~J}$

Option: 3
$5452 \mathrm{~J}$

Option: 4
$21542 \mathrm{~J}$

Answers (1)

best_answer

Given data:

Pressure, $\mathrm{ P=2 \mathrm{~atm}}$

Initial volume, $\mathrm{V_1=0.02 \mathrm{~m}^3}$

Final volume, $\mathrm{V_2=0.04 \mathrm{~m}^3}$

Heat added, $\mathrm{Q=1500 \mathrm{~J}}$

First, calculate the work done

$\mathrm{ W=P \cdot \Delta V }$

$\mathrm{ W=2 \mathrm{~atm} \cdot\left(0.04 \mathrm{~m}^3-0.02 \mathrm{~m}^3\right) \cdot 101325 \mathrm{~Pa} / \mathrm{atm} }$

$\mathrm{ W=20265 \mathrm{~J} }$

Now, use the First Law of Thermodynamics to calculate the change in internal energy:

$\mathrm{\Delta U =Q-W }$

$\mathrm{\Delta U =1500 \mathrm{~J}-20265 \mathrm{~J} }$

$\mathrm{\Delta U =-18765 \mathrm{~J} }$

Hence oprion 1 is correct.

Posted by

manish painkra

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