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Thermodynamics NEET Questions — Free MCQ Practice

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In Column I different Process is given match corresponding option of column II Column - I (A) adiabatic process (B) Isobaric process (C) Isochroic process (D) Isothermal process Column - II (p) $ \triangle p = 0 $ (q) $ \triangle u = 0 $ (.r) $ \triangle Q = 0 $ (s) $ \triangle W = 0 $

A. A-p, B-s, C-r, D-q B. A-s, B-q, C-p, D-r C. A-r, B-p, C-s, D-q D. A-q, B-r, C-q, D-p

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Explanation

adiabatic process $ \triangle Q = 0 $ Isobasic process P = const $ \therefore \triangle P = 0 $ Isochroic process V = const $ \therefore \triangle W = 0 $ Isothermal process T = const $ \therefore \triangle u = 0 $

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In a containes of negligible heat capacity, 200g ice at $0 ^\circ C $ and 100g steam at $ 100 ^\circ C $ are added to 200g of water that has temperature $ 55 ^\circ C $ . Assume no heat is lost to the surroundings and the pressure in the container is constant 1 atm. What is the final temperature the System ?

A. $ 72 ^\circ C $ B. $ 48 ^\circ C $ C. $ 100 ^\circ C $ D. $ 94 ^\circ C $

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Explanation

head required by ice and water to go up to $ 100^\circ C = m_1L+ m_1 sw \triangle T + mw sw \triangle T$ $ = 200 \times 80+200 \times 1 \times 100+200 \times 1 \times 45 $ $= 16,000+20,000+9,000 = 45,000 cal$ $ = give by m_s mass of steam = ms L$ $ ms = { 45,000 \over 540 } = 83.3 g$ $ convert into waters of 100 ^\circ C $ $ Total water = 200 + 200 + 83.3 = 483.3 g$ $ steam left = 100 - 83.3 = 16.79$

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In a containes of negligible heat capacity, 200g ice at $0 ^\circ C $ and 100g steam at $ 100 ^\circ C $ are added to 200g of water that has temperature $ 55 ^\circ C $ . Assume no heat is lost to the surroundings and the pressure in the container is constant 1 atm. At the final temperature, mass of the total water present in the system is

A. 493.6 g B. 483.3 g C. 472.6 g D. 500g

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Explanation

head required by ice and water to go up to $ 100^\circ C = m_1L+ m_1 sw \triangle T + mw sw \triangle T$ $ = 200 \times 80+200 \times 1 \times 100+200 \times 1 \times 45 $ $= 16,000+20,000+9,000= 45,000 cal$ $= give by m_s mass of steam = ms L$ $ ms = { 45,000 \over 540 } = 83.3 g$ $convert into waters of 100 ^\circ C $ $Total water = 200 + 200 + 83.3 = 483.3 g$ $ steam left = 100 - 83.3 = 16.79$

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In a containes of negligible heat capacity, 200g ice at $0 ^\circ C $ and 100g steam at $ 100 ^\circ C $ are added to 200g of water that has temperature $ 55 ^\circ C $ . Assume no heat is lost to the surroundings and the pressure in the container is constant 1 atm. Amount of the Sm left in the system, is equal to

A. 16-7 g B. 8-4g C. 12g D. 0g

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Explanation

head required by ice and water to go up to $ 100^\circ C = m_1L+ m_1 sw \triangle T + mw sw \triangle T$ $ = 200 \times 80+200 \times 1 \times 100+200 \times 1 \times 45 $ $= 16,000+20,000+9,000 = 45,000 cal$ $$= give by m_s mass of steam = ms L$$ $$ ms = { 45,000 \over 540 } = 83.3 g$$ $$convert into waters of 100 ^\circ C $$ $$Total water = 200 + 200 + 83.3 = 483.3 g$$ $$steam left = 100 - 83.3 = 16.79$$

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The temperature of n moles of an ideal gas is increased from $T_{0} t o 2 T_{0}$ through a process $P = \frac{α}{T}$ . Find the work done by the gas. [This question is only for Dropper and XII batch]

A.

$\frac{n R T_{0}}{2}$

B.

$n R T_{0}$

C.

$\frac{3}{2} n R T_{0}$

D.

2 $n R T_{0}$

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Explanation

PV= nRT (ideal gas equation) .....(i)

and $P = \frac{α}{T}$ .....(ii)

Divinding (i) by (ii), we get $V = \frac{n R T^{2}}{α} o r d V = \frac{2 n R T}{α} d T$

$.^{.} . W = \int_{V_{i}}^{V_{f}} P d V = \int_{T_{0}}^{2 T_{0}} (\frac{α}{T}) (\frac{2 n R T}{α}) d T = 2 n R T_{0}$

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A system is taken from state A to state B along two different paths 1 and 2. If the heat absorbed and work done by the system along these two paths are $Q_{1}, Q_{2} a n d W_{1}, W_{2}$ respectively, then

A.

$Q_{1} = Q_{2}$

B.

$W_{1} = W_{2}$

C.

$Q_{1} - W_{1} = Q_{2} - W_{2}$

D.

$Q_{1} + W_{1} = Q_{2} + W_{2}$

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Explanation

Internal energy be state function i.e. not depend the paths. From first law of thermodynamics, Q=U+W

so, $Q_{1} - W = Q_{2} - W$

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The ratio of the relative rise in pressure for adiabatic compression to that for isothermal compression is

A.

$γ$

B.

$\frac{1}{γ}$

C.

1- $γ$

D.

$\frac{1}{1 - γ}$

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Explanation

$PV = k . . . . . . . . Isothermal Process VdP + pdV = 0 {(\frac{dP}{P})}_{Isothermal} = \frac{- dV}{V} {PV}^{γ} = k V^{γ} dP + {γPV}^{γ - 1} dV = 0 . . . . . . . . Adiabatic process {(\frac{dP}{P})}_{Adiabatic} = \frac{- γdV}{V} \frac{{(\frac{dP}{P})}_{Adiabatic}}{{(\frac{dP}{P})}_{Isothermal}} = γ$

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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?

A.

zeroth

B.

First

C.

Second

D.

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

A.

$γ$

B.

$\frac{γ}{γ}$

C.

$1 - \frac{1}{γ}$

D.

$γ - 1$

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Explanation

3.

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

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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

A.

900 J

B.

1800 J

C.

450 J

D.

2700 J

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Explanation

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

Thermodynamics MCQs for NEET — Physics Questions with Answers

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