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Chapter 8: Problem 1210

In Column I different Process is given match corresponding option of column \(\mathrm{I}_{1}\) Column - I Columm - II (A) adiabatic process (p) \(\Delta \mathrm{p}=0\) (B) Isobaric process (q) \(\Delta \mathrm{u}=0\) (C) Isochroic process (r) \(\Delta Q=0\) (D) Isothermal process (s) \(\Delta \mathrm{W}=0\) (A) A-p, B-s, C-r, D-q (B) (B) A-s, B-q, C-p, D-r (C) $\mathrm{A}-\mathrm{r}, \mathrm{B}-\mathrm{p}, \mathrm{C}-\mathrm{s}, \mathrm{D}-\mathrm{q}$ (D) $\mathrm{A}-\mathrm{q}, \mathrm{B}-\mathrm{r}, \mathrm{C}-\mathrm{q}, \mathrm{D}-\mathrm{p}$

Short Answer

Expert verified
The short answer is: (C) A-r, B-p, C-s, D-q.

Step by step solution

01

Understand the four processes

Before we can match the processes with their properties, we need to understand what each process represents: - Adiabatic process: A process in which there is no heat exchange between the system and its surroundings (\( \Delta Q = 0\)). - Isobaric process: A process that occurs at a constant pressure (\( \Delta p = 0\)). - Isochoric process: A process that occurs at a constant volume (\( \Delta W = 0\)). - Isothermal process: A process that occurs at a constant temperature, resulting in no change in internal energy (\( \Delta u = 0\)). 2.
02

Match the processes with their properties

Now we can match each process in Column I to its corresponding property in Column II: - (A) Adiabatic process matches with (r) \( \Delta Q = 0\). - (B) Isobaric process matches with (p) \( \Delta p = 0\). - (C) Isochoric process matches with (s) \( \Delta W = 0\). - (D) Isothermal process matches with (q) \( \Delta u = 0\). 3.
03

Choose the correct option

Based on our matching in step 2, we can see that the correct answer is Option (C): A-r, B-p, C-s, D-q.

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Most popular questions from this chapter

A carnot's engine whose sink is at a temperature of \(300 \mathrm{~K}\) has an efficiency of \(40 \%\) By what amount should the temperature of the source change to increase the efficiency to \(60 \%\) (A) \(275 \mathrm{~K}\) (B) \(325 \mathrm{~K}\) (C) \(300 \mathrm{~K}\) (D) \(250 \mathrm{~K}\)

\(200 \mathrm{~g}\) of water is heated from $25^{\circ} \mathrm{C}^{\circ} 45^{\circ} \mathrm{C}$ Ignoring the slight expansion of the water the change in its internal energy is (Specific heat of wafer \(1\left\\{(\right.\) cal \(\left.) /\left(9^{\circ} \mathrm{C}\right)\right\\}\) (A) \(33.4 \mathrm{KJ}\) (B) \(11.33 \mathrm{KJ}\) (C) \(5.57 \mathrm{KJ}\) (D) \(16.7 \mathrm{KJ}\)

The change in internal energy, when a gas is cooled from $927^{\circ} \mathrm{C}\( to \)27^{\circ} \mathrm{C}$ (A) \(200 \%\) (B) \(100 \%\) (C) \(300 \%\) (D) \(400 \%\)

\(\mathrm{Cp}\) and Cv denote the specific heat of oxygen per unit mass at constant Pressure and volume respectively, then (A) \(\mathrm{cp}-\mathrm{cv}=(\mathrm{R} / 16)\) (B) \(\mathrm{Cp}-\mathrm{Cv}=\mathrm{R}\) (C) \(\mathrm{Cp}-\mathrm{Cv}=32 \mathrm{R}\) (D) \(\mathrm{Cp}-\mathrm{Cv}=(\mathrm{R} / 32)\)

Efficiency of a Carnot engine is \(50 \%\), when temperature of outlet is $500 \mathrm{~K}\(. in order to increase efficiency up to \)60 \%$ keeping temperature of intake the same what is temperature of out let. (A) \(200 \mathrm{~K}\) (B) \(400 \mathrm{~K}\) (C) \(600 \mathrm{~K}\) (D) \(800 \mathrm{~K}\)
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