Two moles of helium gas undergo a cyclic process as shown in figure. Assuming the gas to be ideal, the net work done by the gas is:

200Rln2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

100Rln2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

300Rln2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

400Rln2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D 200Rln2
Work done during process $A \to B (A B)$ (Isobaric process)
$W_{A B} = n R (T_{B} - T_{A})$
$= 100 n R$
Work done during process $B \to C (B C)$ (Isothernal process)
$W_{B C} = n R T_{B} l n (\frac{P_{n}}{P_{c}})$
$= n R T_{B} l n (2)$

Work done during process $C \to D (C D)$
$W_{C D} = n R (T_{C} - T_{D})$
$= - 100 n R$

$W_{n e t} = W_{A B} + W_{D C} + W_{C D} + W_{D A}$

$= n R 300 l n (\frac{1}{2}) + n R 400 l n (2)$

$= 100 n R [3 ln (\frac{1}{2}) + 4 l n (2)]$

$= 100 n R [l n {(\frac{1}{2})}^{3} \times (2)^{4}]$

$= 100 n R l n (2)$

$W_{n e t} = 200 R l n (2)$ $[∵ n = 2]$

Suggest Corrections