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JEE Main 2019 (10 Jan, Shift 2) · Thermodynamic Processes

Thermodynamic Processes question from JEE Main 2019 (10 Jan, Shift 2)

Q217JEE Main 2019 · 10 JAN · SHIFT 2Medium

Thermodynamic Processes

Half mole of an ideal monoatomic gas is heated at constant pressure of 1 atm from 20oC to 90oC. Work done by gas is close to – (Gas constant R = 8.31 J/mol.K)

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Answer: (d)

At constant pressure $W = nR\Delta T$ .
$W = 0.5 \times 8.31 \times (90 - 20) = 0.5 \times 8.31 \times 70 \approx 291\text{ J}$
So the answer is (d) 291 J.

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