Question types

PART - 2 CH - 11 Thermodynamics question types

157 questions across 8 question groups — pick any mix to generate a Physics paper with step-by-step answer keys.

157
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8
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5
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Sample Questions

PART - 2 CH - 11 Thermodynamics questions

One sample from each question group in this chapter. Select any group above to see the full set with answer keys.

If the specific heats per gram of an ideal gas of molecular weight M are represented by Cp and $C _{ \text v }$ then :
  • A
    $C _{ P }- C _{ V }=\frac{ R }{ M ^2}$
  • B
    $C _{ P }- C _{ V }= R$
  • $C_p-C_V=\frac{R}{M}$
  • D
    $C _{ P }- C _{ V }= MR$

Answer: C.

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An engine works between temperatures of $727^{\circ}\text C $ and $227^{\circ}\text C.$ The maximum possible efficiency of the engine is :
  • $\frac{1}{2}$
  • B
    $\frac{1}{4}$
  • C
    $\frac{3}{4}$
  • D
    $1$

Answer: A.

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Which of the following processes is irreversible?
  • A
    Slow isothermal change
  • B
    Slow adiabatic change
  • C
    Evaporation process
  • Diffusion of gases

Answer: D.

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The volume of one litre of gas is constant at 1 Atomos $(10^5 N / m ^2).$ It doubles as it expands under pressure. What will be the work done by the gas?
  • A
    $10^5$ joules
  • B
    $10^4$ joules
  • C
    $10^3$ joules
  • $10^2$ joules

Answer: D.

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Complete conversion of heat into work by a device, it is not possible. This statement is given :
  • A
    From the zeroth law of thermodynamics
  • B
    From the first law of thermodynamics
  • From the second law of thermodynamics
  • D
    According to Joule's law

Answer: C.

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A thermos flask is filled with water. The thermos is shaken for some time due to which the temperature of the water increases. Give reasons. (i) Was work done on water? (ii) Was heat given to water? (iii) Will the internal energy of water increases?
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Three process $A, B, C$ of expansion of a gas from initial volume $V_1$ to final volume $V_2$ are shown in the figure tell which of these change is isothermal, isobaric and adiabatic?
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Q 223 Marks Question3 Marks
As gas expands from initial volume $V_1$ to $V_2$ in three process whose $P$$-V$ curve is shown in the figure, which process is which? In which of the following processes is the gas used the most?
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Q 243 Marks Question3 Marks
The efficiency of an ideal engine is $\frac{1}{8}$ access. After reducing the temperature 100 K , increase to $\frac{1}{6}$ of access. Find the original and final temperature.
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Q 253 Marks Question3 Marks
The output power of the motor of a refrigerator is 240 W . The temperature of the freezing chamber is 240 K and the outside air is 300 K . How much heat can be removed from the freezer in 10 minutes? Assuming that there is ideal efficiency, what is the minimum time in which 10 kg of water at 273 K can be converted in to ice? $J =4.2 \times 10^3 JK cal ^{-1}$.
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0.1 gm molecule of diatomic gas is isothermally expanded at 293 K to twice its initial volume. After that its adiabatic expansion is done until the final volume becomes four times the initial volume. Find the final temperature of the gas and also calculate the work done for both the expansion ( $R = 8 . 3$ joule $m o l ^{ l } K ^{-1}$, $\gamma=1.4)$
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Prove that at any point in the PV diagram, the slope of the adiabatic curve is $\gamma$ times the slope of the isothermal curve. Make necessary diagram.
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AB
1. Efficiency of heat engine $\eta=$(A) $\frac{\Delta P}{\Delta V}$
2. The slope of the adiabatic curve is greater than the isothermal curve(B) Zero
3. Value of W for isochoric process will be(C) $\frac{ Q _2}{ Q _1- Q _2}$
4. The value of performance coefficient $\alpha$ of the refrigerant will be(D) Equal to initial value
5. The working fluid has internal energy after completing the Carnot cycle(E)  $\frac{ W }{ Q }$
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AB
1. A thermos bottle filled with tea is shaken vigorously in the internal energy of the system(A) dQ = 0
2. In adiabatic compression of the body increases(B) will increase
3. For isothermal operation of the engine(C) heat
4. Work done in isothermal process(D) $T _1= T _2$
5. In adiabatic change of gas specific heat value happens zero because(E)$\begin{array}{l} 2.303 P _1 V_1  \log 10 \frac{V_2}{V_1}\end{array}$
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