An ideal gas is taken from point A to the point B , as shown in the p-V diagram, keeping the temperature constant. The work done in the process is :

Question

An ideal gas is taken from point A to the point B , as shown in the p-V diagram, keeping the temperature constant. The work done in the process is : (A)  (pA-pB) (VB-VA)  (B)  (1/2)(pB-pA) (VB+VA)  (C)  (1/2)(pB-pA) (VB-VA)  (D)  (1/2)(pA+pB) (VB+VA)

Tardigrade Admin · Accepted Answer

The p-V diagram is shown below <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/37452c1d1b53cb59e865de092079d5e0-.png" /> Work done = area of ABCDEA = area of Δ ABE + area of rectangle BCDE =(1/2)(pA-pB)(VB-VA)+pB(VB-VA) =[(1/2)(pA-pB)+PB](VB-VA) =(1/2)(pA+pB)(VB-VA)

Q. An ideal gas is taken from point $A$ to the point $B$ , as shown in the $p - V$ diagram, keeping the temperature constant. The work done in the process is :

$(p_{A} - p_{B}) (V_{B} - V_{A})$

$\frac{1}{2} (p_{B} - p_{A}) (V_{B} + V_{A})$

$\frac{1}{2} (p_{B} - p_{A}) (V_{B} - V_{A})$

$\frac{1}{2} (p_{A} + p_{B}) (V_{B} + V_{A})$

The $p - V$ diagram is shown below

Work done = area of $A BC D E A$
= area of $Δ A BE$ + area of rectangle $BC D E$
$= \frac{1}{2} (p_{A} - p_{B}) (V_{B} - V_{A}) + p_{B} (V_{B} - V_{A})$
$= [\frac{1}{2} (p_{A} - p_{B}) + P_{B}] (V_{B} - V_{A})$
$= \frac{1}{2} (p_{A} + p_{B}) (V_{B} - V_{A})$

Q. An ideal gas is taken from point A to the point B , as shown in the p−V diagram, keeping the temperature constant. The work done in the process is :

(pA​−pB​)(VB​−VA​)

21​(pB​−pA​)(VB​+VA​)

21​(pB​−pA​)(VB​−VA​)

21​(pA​+pB​)(VB​+VA​)