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.

Answer 1

Given: Number of moles of the gas, n = 2 moles
The system's volume is constant for lines bc and da. Therefore,
∆V = 0

Thus, work done for paths da and bc is zero.
$\Rightarrow W_{da}=W_{bc}=0$
Since the process is cyclic, ∆U is equal to zero.

Using the first law, we get
∆W = ∆Q
∆W = ∆W_AB + ∆W_CD

Since the temperature is kept constant during lines ab and cd, these are isothermal expansions.
Work done during an isothermal process is given by
W = nRT$\ln \frac{V_f}{V_i}$

If $V_f\mathrm{and}{V}_i$ are the initial and final volumes during the isothermal process, then
$W=n{\mathrm{RT}}_{1}ln\left(\frac{2\mathit{}{\mathrm{V}}_0}{{\mathrm{V}}_0}\right)+n{\mathrm{RT}}_{2}ln\left(\frac{{\mathrm{V}}_0}{2{\mathrm{V}}_0}\right)$
W = nR × 2.303 × log 2 × (500 − 300)
W = 2 × 8.314 × 2.303 × 0.301 × 200
W = 2305.31 J