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NEET
Exam
NEET
Subject
Physics
Chapter
Thermodynamics

NEET · Physics · Thermodynamics

Match of the following. Column I Column II array ll A. Ratio of (Δ...

Match of the following.
<tableclass="qx−match−listqx−match−table"border="1"cellpadding="6"cellspacing="0"><tr><td><br>ColumnI</td><td>ColumnII</td></tr><tr><td><br><table class="qx-match-list qx-match-table" border="1" cellpadding="6" cellspacing="0"><tr><td><br> Column I</td><td>Column II</td></tr><tr><td><br><br>A.RatioofΔQΔUinanisobaric</td></tr><tr><td><br>process<br>\begin{array}{ll} <br> A. Ratio of \frac{\Delta Q}{\Delta U} inan isobaric</td></tr><tr><td><br> process<br> \end{array}</td><td>1.T1(T1−T2)</td></tr><tr><td><br></td><td>1. \frac{T_1}{\left(T_1-T_2\right)}</td></tr><tr><td><br><br>B.RatioofΔQΔWinanisobaric</td></tr><tr><td><br>process<br>\begin{array}{ll} <br> B. Ratio of \frac{\Delta Q}{\Delta W} inan isobaric</td></tr><tr><td><br> process<br> \end{array}</td><td>2.T2(T1−T2)</td></tr><tr><td><br></td><td>2. \frac{T_2}{\left(T_1-T_2\right)}</td></tr><tr><td><br><br>C.Coefficientofperformance</td></tr><tr><td><br>ofarefrigerator<br>\begin{array}{l} <br> C. Coefficient of performance</td></tr><tr><td><br> of a refrigerator<br> \end{array}</td><td>3.γγ−1</td></tr><tr><td><br></td><td>3. \frac{\gamma}{\gamma-1}</td></tr><tr><td><br><br>D.Coefficientofperformance</td></tr><tr><td><br>ofaheatpump<br>\begin{array}{l} <br> D. Coefficient of performance</td></tr><tr><td><br> of a heat pump<br> \end{array}</td><td>4.γ</td></tr><tr><td><br><br></td></tr></table><br>Codes<br></td><td>4. \gamma</td></tr><tr><td><br> <br></td></tr></table><br> Codes<br>ABCD\begin{array}{llll}A & B & C & D\end{array}$$

Options

  1. A4321\begin{array}{llll}4 & 3 & 2 & 1\end{array}
  2. B2143\begin{array}{llll}2 & 1 & 4 & 3\end{array}
  3. C3124\begin{array}{llll}3 & 1 & 2 & 4\end{array}
  4. D4231\begin{array}{llll}4 & 2 & 3 & 1\end{array}

Correct answer

A. 4321\begin{array}{llll}4 & 3 & 2 & 1\end{array}

Step-by-step solution

We know that, at constant volume, Heat supplied, ΔQ1=ΔU=CVΔT\Delta Q_1=\Delta U=C_V \Delta T At constant pressure, heat supplied, ΔQ=CpΔT=ΔU+ΔW\Delta Q=C_p \Delta T=\Delta U+\Delta W Ratio of heat capacity at constant pressure to constant volume is γ=CpCV\gamma=\frac{C_p}{C_V} Now, ΔQΔU=CpΔTCVΔT=CpCV=γ\frac{\Delta Q}{\Delta U}=\frac{C_p \Delta T}{C_V \Delta T}=\frac{C_p}{C_V}=\gamma Now, ΔQΔW=ΔQΔQ−ΔU=CpΔTCpΔT−CVΔT\frac{\Delta Q}{\Delta W}=\frac{\Delta Q}{\Delta Q-\Delta U} =\frac{C_p \Delta T}{C_p \Delta T-C_V \Delta T}=CpCp−CV=CpCV(CpCV−1)=γγ−1\begin{aligned} & =\frac{C_p}{C_p-C_V}=\frac{C_p}{C_V\left(\frac{C_p}{C_V}-1\right)} \\ & =\frac{\gamma}{\gamma-1} \end{aligned}Forrefrigerator,coefficientofperformance,For refrigerator, coefficient of performance,\begin{aligned} \beta & =\frac{Q_2}{W}=\frac{\text{ Heat absorbed from food stuff }}{\text{ Electrical work done }} \\ \therefore \quad \beta & =\frac{Q_2}{Q_1-Q_2} \end{aligned}$CoefficientofheatpumpCoefficient of heat pump=\frac{\text{ Heat released }}{\text{ Electrical work }}i.e.i.e.\quad \alpha=\frac{Q_1}{W}=\frac{Q_1}{Q_1-Q_2}Weknowthat,We know that,\frac{Q_1}{Q_2}=\frac{T_1}{T_2}So,weget,So, we get,\beta=\frac{Q_2}{Q_1-Q_2}=\frac{\frac{Q_2}{Q_1}}{1-\frac{Q_2}{Q_1}} =\frac{\frac{T_2}{T_1}}{1-\frac{T_2}{T_1}}=\frac{T_2}{T_1-T_2}andand\alpha=\frac{Q_1}{Q_1-Q_2} =\frac{\frac{Q_1}{Q_2}}{\frac{Q_1}{Q_2}-1}=\frac{\frac{T_1}{T_2}}{\frac{T_1}{T_2}-1} =\frac{T_1}{T_1-T_2}Hence,wegetthebestmatchasgivenbyHence, we get the best match as given by\mathrm{A} \rightarrow 4, \mathrm{ B} \rightarrow 3, \mathrm{C} \rightarrow 2andand\mathrm{D} \rightarrow 1$.
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