An ideal gas is taken around the cycle, ABCA as shown in P-V diagram. The net work done during the cycle is equal to

Question

An ideal gas is taken around the cycle, ABCA as shown in P-V diagram. The net work done during the cycle is equal to <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/c-qw8dvayxse4g.png" /> (A) 12P1V1 (B) 6P1V1 (C) 5P1V1 (D) P1V1

Tardigrade Admin · Accepted Answer

Work done = Area bounded by the triangle ABC =(1/2)× 2V1× 5P1 =5P1V1

Q. An ideal gas is taken around the cycle, ABCA as shown in P-V diagram. The net work done during the cycle is equal to

$12 P_{1} V_{1}$

$6 P_{1} V_{1}$

$5 P_{1} V_{1}$

$P_{1} V_{1}$

Work done = Area bounded by the triangle ABC

$= \frac{1}{2} \times 2 V_{1} \times 5 P_{1}$

$= 5 P_{1} V_{1}$

Q. An ideal gas is taken around the cycle, ABCA as shown in P-V diagram. The net work done during the cycle is equal to

12P1​V1​

6P1​V1​

5P1​V1​

P1​V1​

Work done = Area bounded by the triangle ABC =21​×2V1​×5P1​ =5P1​V1​

$12 P_{1} V_{1}$

$6 P_{1} V_{1}$

$5 P_{1} V_{1}$

$P_{1} V_{1}$

Work done = Area bounded by the triangle ABC

$= \frac{1}{2} \times 2 V_{1} \times 5 P_{1}$

$= 5 P_{1} V_{1}$