NEET-XII-Physics
26: Laws of Thermodynamics
- #18Figure 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.
FigureAns : 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 = ∆WAB + ∆WCD
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`` \,\mathrm{\,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{\,\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
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