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 K, ln2 = 0.6931] (Round off to the nearest integer) Numerical answer type. Work it out, then open the answer.

Answer: 17258. PAVA = 200 × 2 = 400 and PBVB = 100 × 4 = 400, so the process is isothermal. W = nRTln (V₂)/(V₁) = 1 × 8.3 × 300 × ln 2 W = 2490 × 0.6931 = 1725.8 J = 17258 × 10⁻¹ J So the answer is 17258.

One mole of an ideal gas at 27 ° C is taken from A to B as shown in… | JEE Main 2021 (20 Jul, Shift 2)

Q51JEE Main 2021 · 20 JUL · SHIFT 2Medium

P-V Diagrams & Cyclic Processes

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

Numerical answer type. Work it out, then open the answer.

Show answer and solution

Answer: 17258

$P_AV_A = 200 \times 2 = 400$ and $P_BV_B = 100 \times 4 = 400$ , so the process is isothermal.
$W = nRT\ln\frac{V_2}{V_1} = 1 \times 8.3 \times 300 \times \ln 2$
$W = 2490 \times 0.6931 = 1725.8\ \text{J} = 17258 \times 10^{-1}\ \text{J}$
So the answer is 17258.

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