Question

Figure shows the variation of internal energy (U) with the pressure (P) of 2.0 mole gas in cyclic process abcda. The temperature of gas at c and d are 300 and 500 K. The heat absorbed by the gas during the process is given by K(100)R ln 2. Find the value of K.

Answer 1

Change in internal energy for cyclic process $\left(\Delta U\right)=0$

for process $a\to b$, (P = constant)

$W_{a\to b}=P\Delta V=nR\Delta T=-400R$

for process $b\to c$, (T = constant)

$W_{b\to c}$ = -2R (300) ln 2

for process c $\to$ d (P = constant)

$W_{c\to d}=+400R$

for process d $\to$ a, (T = constant)

$W_{d\to a}$ = 2R (500) ln 2

$\Delta W=W_{a\to b}+W_{b\to c}+W_{c\to d}+W_{d\to a}$

$\Delta W=400R$ ln 2

$\Delta Q=\Delta W$

$\Delta Q=400R$ln 2 = 4(100) R ln 2

$\therefore$ K = 4