One mole of an ideal gas at 27 °C is taken from A to B as shown in the given PV indicator diagram. The work done by the system will be _______ ×10^(-1) J. [Given, R=8.3 J mole^(-1) K, ln2=0.6931] (Round off to the nearest integer) (Round off to the nearest integer)

Correct answer: 17258. Process of isothermal W=nRTℓnV_2 V_1 =1×8.3×300×ln2 =17258×10^(-1) J

Q87 / 90 · JEE Main 2021 (20 Jul Shift 2) · Physics

One mole of an ideal gas at $27 ° C$ is taken from $A$ to $B$ as shown in the given $P V$ indicator diagram. The work done by the system will be _______ $\times 10^{- 1} J .$ [Given, $R = 8.3 J {mole}^{- 1} K, \ln 2 = 0.6931$ ] (Round off to the nearest integer)
(Round off to the nearest integer)

Correct answer: 17258

Process of isothermal

$W = n R T ℓ n (\frac{V_{2}}{V_{1}})$

$= 1 \times 8.3 \times 300 \times \ln 2$

$= 17258 \times 10^{- 1} J$

Hence, the correct answer is 17258.