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.

#### Solution

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.

⇒ 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} and V_{i} are the initial and final volumes during the isothermal process, then

\[W = n {RT}_1 ln\left( \frac{2 V_0}{V_0} \right) + n {RT}_2 ln\left( \frac{V_0}{2 V_0} \right)\]

W = nR × 2.303 × log 2 × (500 − 300)

W = 2 × 8.314 × 2.303 × 0.301 × 200

W = 2305.31 J