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A sink, that is the system where heat is rejected, is essential for the conversion of heat into work. From which law the above inference follows?

zeroth

First

Second

Third

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Explanation

$I I^{n d}$ law of thermodynamics

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An ideal gas with adiabatic exponent y is heated at constant pressure and it absorbs Q heat. What fraction of this heat is used to perform external work

$γ$

$\frac{γ}{γ}$

$1 - \frac{1}{γ}$

$γ - 1$

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Explanation

$\begin{aligned} \frac{d U}{d Q} = \frac{1}{γ} \\ \frac{d Q - d W}{d Q} = \frac{1}{γ} \\ \frac{d W}{d Q} = (1 - \frac{1}{γ}) \end{aligned}$

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A Carnot engine working between 400K and 800K has a work output of 900J per cycle. The amount of heat energy supplied to engine from the source per cycle is

900 J

1800 J

450 J

2700 J

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Explanation

$(2) 1 - \frac{400}{800} = \frac{900}{Q_{1}}$

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Temperature is defined by

First Law of thermodynamics

Second Law of thermodynamics

Third Law of thermodynamics

Zeroth Law of thermodynamics

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Explanation

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If 32 gm of $O_{2}$ at $27 ° C$ is mixed with 64 gm of $O_{2}$ at $327 ° C$ in an adiabatic vessel, then the final temperature of the mixture will be :

$200 ° C$

$227 ° C$

$314.5 ° C$

$235.5 ° C$

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Explanation

$\frac{n_{1} f R T_{1}}{2} + \frac{n_{2} f R T_{2}}{2} = \frac{(n_{1} + n_{2}) f R T_{3}}{2} \frac{1 \times 300}{2} + \frac{2 \times 600}{2} = \frac{3 \times T}{2} T = 500 K (227 {}_{o}C)$

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If $W_{1}$ is the work done in compressing an ideal gas from a given initial state through a certain volume isothermally and $W_{2}$ is the work done in compressing the same gas from the same initial state through the same volume adiabatically, then:

$W_{1} = W_{2}$

$W_{1} < W_{2}$

$W_{1} > W_{2}$

$W_{1} = 2 W_{2}$

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Explanation

For an ideal gas, the work done in an isothermal compression (Wâ‚) is less than the work done in an adiabatic compression (Wâ‚‚) for the same volume change. This is because in the adiabatic case, the gas also gains internal energy due to the work done against intermolecular forces, making Wâ‚‚ greater than Wâ‚.

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During an experiment and ideal gas is found to obey an additional law $V P^{2}$ =constant. The gas is initially at a temperature T and volume V. When it expand to a volume 2V, the temperature becomes.

T $\sqrt{2}$

T/2

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Explanation

$V p^{2} = c o n s \tan t V \frac{n^{2} R^{2} T^{2}}{V^{2}} = C o n s \tan t (U \sin g I d e a l G a s E q n) T^{2} α V$

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The temperature inside a refrigerator is ${t_{2}}^{\circ} C$ and the room temperature is ${t_{1}}^{\circ} C$ . The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be -

$\frac{t_{1}}{t_{1} - t_{2}}$

$\frac{t_{1} + 273}{t_{1} - t_{2}}$

$\frac{t_{2} + 273}{t_{1} - t_{2}}$

$\frac{t_{1} + t_{2}}{t_{1} + 273}$

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Explanation

(b) For a refrigerator, we know that

$\frac{Q_{1}}{W} = \frac{Q_{1}}{Q_{1} - Q_{2}} = \frac{T_{1}}{T_{1} - T_{2}}$

where,

$Q_{1}$ =amount of heat delivered to the room

W = electrical energy consumed

$T_{1}$ = room temperature= $t_{1} + 273$

$T_{2} =$ temperature of sink= $t_{2} + 273$

$∴$ $\frac{Q_{1}}{1} = \frac{t_{1} + 273}{t_{1} + 273 - (t_{2} + 273)}$

$\Rightarrow Q_{1} = \frac{t_{1} + 273}{t_{1} - t_{2}}$

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A gas is compressed isothermally to half its initial volume. The same gas is compressed separately through an adiabatic process until its volume is again reduced to half .Then -

compressing the gas through adiabatic process will require more work to be done

compressing the gas isothermally or adiabatically will require the same amount work

which of the case (whether compression through isothermal or through adiabatic process) requires more work will depend upon the atomicity of the gas

compressing the gas isothermally will require more work to be done

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A refrigerator works between 4°C and 30°C. It is required to remove 600 calories of heat every second in order to keep the temperature of the refrigerated space constant. The power required is (Take, 1 cal = 4.2 Joules)

23.65W

236.5W

2365W

2.365W

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Explanation

(b) Given temperature of source T=30°C=30+273 T1=303K
Temperature of sink T₂=4°C=4+273 T₂=277K

As we know that
Q₁/Q₂=T₁/T₂
=>Q₂+W/Q₂=T₁/T₂ ..........(W=Q₁-Q₂)

where Q₂ is the amount of heat drawn from the sink (at T₂),W is workdone on working substance,
Q₁ is amount of heat rejected to source (at room temperature $T_{1}$ ).

=>WT₂+T₂Q₂=T₁Q₂
=>WT₂=T₁Q₂-T₂Q₂
=>WT₂=Q₂(T₁-T₂)
=>W=Q₂(T₁/T₂-1)
=>W=600X4.2X(303/277-1)
W=600X4.2X(26/277)
W=236.5Joules

Power=Workdone/Time=W/t=236.5/1=236.5W

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