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Physics MCQs for NEET — Practice Questions with Answers

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At constant temperature, if pressure increases by 1%, the percentage decrease of volume is

100/101%

1/101%

1/100%

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Explanation

$P_{1} V_{1} = P_{2} V_{2}$

If P₁ = 100 mm, P₂ will be 101 mm

Hence $100 \times V = 101 \times V_{2}$ ,

$V_{2} = \frac{100}{101} \times V$ ,

Decrease in volume $= V - \frac{100 V}{101} = \frac{1}{101}$ of V i.e. $\frac{100}{101} %$

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When the temperature of 23 ml of dry CO₂ gas is changed from 10° to 30°C at constant pressure of 760 mm, the volume of gas becomes closest to which one of the following [CPMT 1992]

7.7 ml

25.5 ml

24.6 ml

69 ml

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Explanation

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ i.e. $\frac{23}{283 K} = \frac{V_{2}}{303 K}$ ; $V_{2} = 24 .6 m l$

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The volume of a gas is 100 ml at 100°C. If pressure remains constant then at what temperature it will be about 200 ml [Roorkee 1993]

200°C

473°C

746°C

50°C

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Explanation

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ i.e. $\frac{100}{373 K} = \frac{200}{T_{2}}$

$T_{2} = 746 k = 473^{o} C$

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If 300 ml of a gas at 27°C is cooled to 7°C at constant pressure, its final volume will be [AIIMS 2000]

135 ml

540 ml

350 ml

280 ml

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Explanation

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ i.e. $\frac{300}{300 K} = \frac{V_{2}}{280 K}$

V₂ = 280 ml

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A flask containing air (open to atmosphere) is heated from 300 K to 500 K. the percentage of air escaped to the atmosphere is nearly

16.6

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Explanation

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ i.e. $\frac{V_{1}}{300} = \frac{V_{2}}{500}$ ; $V_{2} = \frac{5}{3} V = 1 .66 V$

Volume escaped $= 1 .66 V - V = 0 .66 V = 66 %$ of V

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According to Charle’s law, at constant pressure, 100 ml of a given mass of a gas with 10°C rise in temperature will become $(\frac{1}{273} = 0 .00366)$

100.0366

99.9634

103.66

100.366

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A sealed tube which can withstand a pressure of 3 atmosphere is filled with air at $27^{o} C$ and 760 mm pressure. The temperature above which the tube will burst will be

900°C

627°C

726°C

1173°C

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Explanation

The tube will burst when the final pressure exceeds 3 atm. at constant volume,

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$ i.e. $\frac{760}{300 K} = \frac{3 \times 760}{T_{2}}$

$T_{2} = 900 K = 627^{o} C$

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The pressure of 2 moles of an ideal gas at 546 K having volume 44.8 L is [CPMT 1995]

2 atm

3 atm

4 atm

1 atm

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Explanation

$P V = n R T, P \times 44 .8 = 2 \times 0 .082 \times 546$

P = 2 atm

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The number of moles of H₂ in 0.224 litre of hydrogen gas at STP (273 K, 1 atm.) is [MLNR 1994]

0.1

0.01

0.001

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Explanation

$P V = n R T$ , $1 \times 0 .224 = n \times 0 .082 \times 273 n = 0 .01 m o l$

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120 g of an ideal gas of molecular weight 40 mole^–1 are confined to a volume of 20 L at 400 K. Using $R = 0 .0821 L a t m K^{- 1} m o l e^{- 1}$ , the pressure of the gas is [Pb. CET 1996]

4.90 atm

4.92 atm

5.02 atm

4.96 atm

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Explanation

$120 g = \frac{120}{40} = 3 m o l e s$

$P = \frac{n R T}{V} = \frac{3 \times 0 .0821 \times 400}{20} = 4 .92 a t m .$

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