NEET-XII-Physics
26: Laws of Thermodynamics
- #11A gas is taken through a cyclic process ABCA as shown in figure. If 2.4 cal of heat is given in the process, what is the value of J ?
FigureAns : Heat given in the process, ∆Q = 2.4 cal
∆W = WAB + WBC + WCA
For line AB:
Change in volume, `` ∆V=0``
`` \therefore {W}_{\,\mathrm{\,AB\,}}=0``
For line BC:
`` \,\mathrm{\,Mean\,}\,\mathrm{\,pressure\,},P=\frac{1}{2}\times (100+200)\,\mathrm{\,kPa\,}=150\times {10}^{3}\,\mathrm{\,Pa\,}``
`` ∆V=(700-500)\,\mathrm{\,cc\,}=200\,\mathrm{\,cc\,}``
`` {W}_{\,\mathrm{\,BC\,}}=150\times {10}^{3}\times 200\times {10}^{-6}``
`` {W}_{\,\mathrm{\,BC\,}}=30\,\mathrm{\,J\,}``
For line AC:
P = 100 kPa
`` ∆V=200cc``
WCA = 100 × 103 × 200 × 10-6
Total work done in the one cycle is given by
∆W = 0 + 150× 103 × 200 × 10-6 - 100 × 103 × 200 × 10-6
∆W = `` \frac{1}{2}`` × 300 × 103 × 200 × 10-6 - 20
∆W = 30 J - 20 J = 10 J
∆U = 0 (in a cyclic process)
∆Q = ∆U + ∆W
⇒ 2.4J = 10
`` \Rightarrow J=\frac{10}{2.4}=\frac{100}{24}=\frac{25}{6}=4.17\,\mathrm{\,J\,}/\,\mathrm{\,cal\,}``
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