Question

The figure shows the variation of internal energy (U) with the pressure (P) of 2.0 mole gas in cyclic process abcda. The temperature of the gas at c and d are 300 K and 500 K. Calculate the heat absorbed by the gas during the process.

• A. 400 R ln 2
• B. 200 R ln 2
• C. 100 R ln 2
• D. 300 R ln 2

Answer 1

The correct option is A 400 R ln 2

Change in internal energy for cyclic process $\Delta U$ = 0.
For process $a\to b$, P is constant
$W_{a\to b}=P\Delta V$
$=nR\Delta T$
$=-400R$
For process $b\to c$, T is constant
$W_{b\to c}=-2R\left(300\right)ln2$
For process $c\to d$, P is constant
$W_{c\to d}=+400R$
For process $d\to a$ T is constant
$W_{d\to a}=+2R\left(500\right)ln2$
Net Work $\Delta W$ = $W_{a\to b}+W_{b\to c}+W_{c\to d}+W_{d\to a}$
$\Delta W=400Rln2$
$\therefore dQ=dU+dW$, first law of thermodynamics
$\therefore dQ=400Rln2$