Question

Figure shows the variation in the internal energy U with the volume V of 2.0 mol of an ideal gas in a cyclic process abcda. The temperatures of the gas at b and c are 500 K and 300 K respectively. Calculate the heat absorbed by the gas during the process.
Figure

Answer 1

The heat absorbed by the gas during the process is 2305.31 J .

Explanation:

Step 1:

n = 2 moles

n is Number of moles of the gas

The volume of the system for the lines bc and da is constant. Accordingly,

∆V = 0

Step 2:

Therefore, work done for the da and bc paths is zero.

$\begin{aligned}&\mathrm{W}_{\mathrm{d}_{a}=\mathrm{W}_{\mathrm{b}_{\mathrm{c}}}=0\\&W_{d{a}}=W_{bc}=d\end{aligned}$

because the process is cyclic, ∆U = zero.

∆W = ∆Q

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

These are isothermal expansions, since the temperature is held constant during lines ab and cd.  

Step 3:

Work is performed during an isothermal process

$\mathrm{W}=\mathrm{nRT} \ln \frac{\mathrm{v} f}{\mathrm{v}_{i}}$

The original and the final volumes during the isothermal process are $V_f$ and $V_i$, then

$\begin{aligned}&W=n R T_{1} \ln \left(\frac{2 V_{o}}{V_{o}}\right)+n R T_{2} \ln \left(\frac{V_{o}}{2 V_{o}}\right)\\&W=\mathrm{n} R \times 2.303 \times \log 2 \times(500-300)\end{aligned}$

W = 2 × 8.314 × 2.303 × 0.301 × 200

W = 2305.31 J

Answer 2

$\huge{\boxed{\mathcal\pink{\fcolorbox{red}{yellow}{Answer}}}}$

2305.31 joule

will be the answer