Question

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

• A. 200Rln2
• B. 100Rln2
• C. 300Rln2
• D. 400Rln2

Answer 1

Answer: (A)

200Rln2

Answer 2

: At constant pressure

$\Delta \mathrm { W } = \mathrm { P } \Delta \mathrm { V } = \mathrm { P } \times \left( \mathrm { V } _ { \mathrm { f } } - \mathrm { V } _ { \mathrm { i } } \right)$ ….. (i)

For an idea gas $P V = n R T$

Or $P V _ { f } = n R T$ and $P V _ { i } = n R T _ { i }$

Form (i)

$\therefore \Delta W = n R \left( T _ { f } - T _ { i } \right) \ldots . . ( i i )$

For constant temperature, $P V =$ constant

$\therefore \Delta W = n R T \ln \frac { P _ { i } } { P _ { f } }$…… (iii)

So work done for path AB, BC, CD and DA

respectively will be

$\Delta W _ { A B } = n R \left( T _ { f } - T _ { i } \right) = 2 \times R ( 400 - 300 ) = 200 R$(using (ii))

(using (iii))

$\Delta W _ { C D } = n R \left( T _ { f } - T _ { i } \right) = 2 \times R [ 300 - 400 ] = - 200 R$

(using (ii))

$\Delta W _ { D A } = n R T \ln \left( \frac { P _ { i } } { P _ { f } } \right) = 2 \times R \times 300 \ln \left( \frac { 1 } { 2 } \right) = - 600 R \ln 2$(using (iii))

Hence, the work done in the complete cycle

$\Delta W = W _ { A B } + W _ { B C } + W _ { C D } + W _ { D A }$

$= 200 R + 800 R \ln 2 - 200 R - 600 R \ln 2 = 200 R \ln 2$