Question

Figure shows the variation of internal energy 'U' with the density of one mole of monoatomic gas for a thermodynamic cycle ABCA. AB process is a rectangular hyperbola. The amount of work done in the process A ® B is :

• A. $\frac { 2 \mathrm { U } _ { 0 } } { 3 }$
• B. –$\frac { 2 \mathrm { U } _ { 0 } } { 3 }$
• C.
• D. –

Answer 1

Answer: (D)

–

Answer 2

For the process A ® B

Ua$\frac { 3 } { 2 }$RT a $\frac { \frac { 1 } { \mathrm { M } } } { \mathrm { v } }$

Or T a V

So AB is isobaric process (Pressure is constant)

T_A = $\frac { 8 \mathrm { U } _ { 0 } } { 3 \mathrm { R } }$ [As 4U₀ = 3/2 RT_A]

V_A = $\frac { \mathrm { M } } { 2 \rho _ { 0 } }$

So p_A =$\frac { 16 \mathrm { U } _ { 0 } \rho _ { 0 } } { 3 \mathrm { M } }$& V_B =

So W_AB = p[v_B – v_A]

̃ $\frac { 16 \mathrm { U } _ { 0 } \rho _ { 0 } } { 3 \mathrm { M } }$ $\left[ \frac { \mathrm { M } } { 4 \rho _ { 0 } } - \frac { \mathrm { M } } { 2 \rho _ { 0 } } \right]$ = – $\frac { 16 \mathrm { U } _ { 0 } \rho _ { 0 } } { 3 \mathrm { M } }$ ×

̃ –