Half mole of an ideal |KTG & THERMODYNAMICS|JEE MAIN 2019

JEE MAIN 2019

10-01-2019 S2

Question

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

Select the correct option:

291 J

581 J

146 J

73 J

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

$\begin{aligned} \Delta \mathrm{W}=\mathrm{P} & \Delta \mathrm{V} \\ & =\mathrm{nR} \Delta \mathrm{T} \\ & =\frac{1}{2} \times 8.31(70) \\ & =291 \mathrm{~J}\end{aligned}$

Question Tags

JEE Main

Physics

Medium

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