Two moles of an ideal gas is changed from its initial state ( $16 a t m, 6 L$ ) to final state ( $4 a t m, 15 L$ ) in such a way that this change can be represented by a straight line in $P - V$ curve. The maximum temperature attained by the gas during the above change is:
(Take $R = \frac{1}{12} L a t m K^{- 1} {m o l}^{- 1}$ )

Question

Two moles of an ideal gas is changed from its initial state (

16 a t m, 6 L

) to final state (

4 a t m, 15 L

) in such a way that this change can be represented by a straight line in

P - V

curve. The maximum temperature attained by the gas during the above change is:

(Take

R = \frac{1}{12} L a t m K^{- 1} {m o l}^{- 1}

)

Toppr Admin · Accepted Answer

Given P1-16 atmP2-4 atmV1-6 LV2-15 LThe shaded portion of the above graph represents the work doneArea-area of the triangle- ABC-Area of rectagleBCED-12-xD7-9-xD7-12-9-xD7-4-90LAtm-WW-2-303-xD7-nRTlogV2V1-90-2-303-xD7-2-xD7-0-083-xD7-T-xD7-log156T-604 K

Given P1=16 atmP2=4 atmV1=6 LV2=15 LThe shaded portion of the above graph represents the work doneArea=area of the triangle= ABC+Area of rectagleBCED=12×9×12+9×4=90LAtm=WW=2.303×nRTlogV2V1-90=2.303×2×0.083×T×log156T=604 K