# ICSE Class 9th Chemistry 7 – Study of Gas laws MCQs

#### The ratio of partial pressures of gaseous components to the total pressure of the mixture represents ____.

#### A balloon has a volume of 200 dm3 when filled with hydrogen gas at a pressure of 1.0 bar. If it rises to a height where the atmospheric pressure is 0.5 bar, the volume of the balloon will be ____. [Assume that temperature is constant].

A balloon has a volume of 200 dm3 when filled with hydrogen gas at a pressure of 1.0 bar. If it rises to a height where the atmospheric pressure is 0.5 bars, the volume of the balloon will be 400 dm^{3}. Using Boyle's law
P_{1}V_{1 }= P_{2}V_{2}
P_{1 }= 1 bar, P_{2} = 0.5 bar, V_{1} = 200 dm_{3}
V_{2 }= P_{1}V_{1}/P_{2} = 1 × 200/0.5 = 400 dm_{3}

#### Pick the wrong statement/statement(s) in the following options

The statement B is wrong since, the size of the degree on Centigrade scale of temperature is equal to the size of the degree on the absolute scale of temperature, since the change in temperature recorded on the Centigrade scale is the same as that recorded on the absolute scale of temperature.

#### 1 mol gas A 2 mol gas B Partition removed at temperature T → Pressure P250 mm 250 mm In the above experiment, the value of P is ____ .

The value of P in the experiment is 250 mm. Let volume of 1 mol of gas A = V Therefore, volume of 2 mol of gas B = 2V (temperature is same) After partition is removed, Total volume = 3V PA = 250 × V /3V = 250/3 PB = 250 × 2V / 3V = 500/3 P = P_{A} + PB = 250 / 3 + 500 / 3 = 250 mm

#### Statement A: Gases deviate from their ideal behaviour at low temperatures. Statement B: At low temperatures, the molecular attractions become significant, leading to a deviation from the ideal behaviour for a gas.

At high temperatures, the molecular motion readily overcomes the molecular attractions and the molecules move about as though there were no molecular attraction. But at low temperatures the molecular attractions become significant and this leads to a deviation from the ideal behaviour for a gas. Hence A and B are true.

#### To completely burn 1 kg of sulphur (S_{8})which contains 3% material which does not burn, ____ volume of air containing 11% of oxygen by volume is required at STP. (Sulphur burns according to the reaction - 1/8 S_{8} + O^{2} ---------------> SO^{2})

The amount of incombustible material = 1000 x 3/100 = 30 g Therefore, mass of pure sulphur = 1000 - 30 = 970 g From equation, we get 32 g of sulphur requires oxygen = 22.4 L at STP Therefore, 970 g of sulphur requires 22.4 x 970 / 32 = 679 L of oxygen. Now, If 11 L of oxygen is needed, the air required = 100 L If 679 L of oxygen is needed, the air required = 679 x 100/ 11 = 6172.73 L Hence, to completely burn 1 kg of sulphur (S8) which contains 3% material which does not burn, 6172.73 L of air containing 11% of oxygen by volume is required at STP.

#### At STP, a container has 1 mole of Argon, 2 moles of carbon dioxide, 3 moles of oxygen and 4 moles of nitrogen. Without changing the total pressure if one mole of oxygen is removed. The partial pressure of oxygen ____.

The partial pressure of oxygen changes by 26%. Total moles of gases = 10. Let the pressure be P. Therefore, partial pressure of oxygen PO2 =3P / 10 No. of moles of oxygen in the mixture, after removal of 1 mol of oxygen = 2 Total moles = 9 Partial pressure of oxygen after removal of oxygen =2P / 9 Decrease in partial pressure of oxygen = 3P / 10 - 2P / 9 = 0.078 P % decrease = 0.078 P x 100 / 3P = 26%

#### A glass bulb of volume 25 cc is filled with oxygen at a pressure of 8.31 × 105 pa and at 300 K. The number of molecules of oxygen in the bulb is ____.

A glass bulb of volume 25 cc is filled with oxygen at a pressure of 8.31 × 10^{5}pa and at 300 K. The number of molecules of oxygen in the bulb is 5 × 10^{21}. Given temperature of oxygen, T = 300 K Volume of oxygen, V = 25 × 10^{-6}m^{-3}
According to ideal gas equation,
PV = µRT(R is universal gas constant, µ is the number of moles of the gas) 8.31 × 10^{5} × 25 × 10^{-6}= µ × 8.31 × 300 Solving we get, number of moles of oxygen, μ=1/120 We know that, one mole of any gas contains 6.023 × 10^{23}molecules. Therefore, the number of molecules of oxygen = µ × 6.023 × 10^{23} =1/120 × 6. 023 × 10^{23}=0. 05×10^{23}=5 × 10^{21}

#### The Gay Lussac's equation is ____.

Gay Lussac's equation is P α T.
Gay Lussac's law states that "the pressure of a fixed mass of a gas is directly proportional to its temperature at constant volume". Thus, P_{1}/T_{1} = P_{2}/T_{2}.

#### 10 L of helium is contained in a balloon at a pressure of 2 bar and 27° C and some oxygen is introduced into the balloon making the final volume as 30 L. The pressure falls to 1 bar. Temperature, however it is kept constant. Then, the partial pressure of oxygen will be ____.

The partial pressure of oxygen will be 0.2 bar. Mol of Helium present = 2 × 10/ R × 300 Partial pressure of helium = 2 × 10 × R × 300/ R × 300 × 25 = 0.8 bar Partial pressure of oxygen = 1 - 0.8 = 0.2 bar

#### A graph is plotted with (PV) on y-axis and V on x-axis where P and V denote the pressure and volume of a fixed mass of gas respectively. At constant temperature, the graph is a ____.

At constant temperature, we know that PV = k where k is a constant for a fixed mass of a gas. Since (PV) = k, this implies that the graph represents a straight line, y = k, which is a straight line parallel to the x-axis. Hence option B is correct

#### It is desired to increase of the volume of a gas by 30% without changing the pressure. If the initial temperature of the gas is 27^{o} C, it must be heated to ____.

The gas must be heated to 117^{o} C.
Let volume of gas at 27^{o} C be V cm^{3}
Desired increase in volume = 30% of V = 30/100 × V = 0.3 V
Therefore, final volume = V + 0.3 V = 1.3 V
V_{1} = V cm_{3}, T_{1} = 300 K,
V sub>2 = 1.3 V
At constant P,
V_{1}/T_{1} = V_{2}/T_{2}
V/300 = 1.3/ T_{2}
T_{2} = 390 K
= 390 - 273 = 117^{o} C

#### One litre of oxygen, hydrogen, nitrogen and carbon-dioxide are considered at S. T. P. The gas which has greatest mass is ____.

One litre of oxygen, hydrogen, nitrogen and carbon-dioxide are considered at S.T.P. The gas which has greatest mass is carbon-dioxide. Given, that all the gases occupy equal volume V, at S.T.P. Hence they are at same absolute temperature T and pressure P. According to ideal gas equation, PV = µRT (R is universal gas constant, µ is the number of moles of the gas) Here μ=mM Where m and M are the mass and molecular weight of the gas respectively. PV=mMRT PVM = mRT Here, P, V, R and T are constants. m∝M Hence, the gas which has the largest molecular weight M, has the largest mass ‘m’. Among oxygen, hydrogen, nitrogen and carbon-dioxide, the gas having greatest molecular weight is carbon-dioxide. Hence carbon-dioxide has the largest mass among the given options.

#### Two glass bulbs A of 100 mL and B of 150 mL capacity containing same gas are connected by a small tube of negligible volume. At a particular temperature, the pressure in A was found to be 20 times more than that in bulb B. The stop cock is opened without changing the temperature. The pressure in A will ____.

The pressure in A will drop by 57%. Let original pressure in B be P. Then, pressure in A will be 20P. After opening the stopcock, the total volume = 250 mL Partial pressure of gas in A be PA: PA × 250 = 100 × 20 P PA = 100×20P/250 = 8 P Let Partial pressure of gas in B be PB: PB × 250 = 150 × P PB = 150P/250 = 0.6P Total pressure = 8P + 0.6P = 8.4P Now drop in pressure in A = 20P - 8.6P = 11.4 P % drop in pressure = 11.4 P/20 P × 100 = 57 %

#### ____ law is called Amonton's law.

Gay Lussac's law is called Amonton's law.
It states that "volume remaining constant, the pressure of a given mass of a gas increases or decreases by 1/273 of its pressure at 0^{o} C, for every 1^{o }C rise or fall in temperature".

#### ____ is kept constant according to Boyle's law, Charle's law and Gay Lussac's law respectively.

Temperature is kept constant according to Boyle's law, pressure is kept constant according to Charle's law and volume is kept constant according to Gay Lussac's law. According to Boyle's law “the volume of a fixed mass of a gas is inversely proportional to the pressure at constant temperature". According to Charle's law “the volume of a fixed mass of a gas is directly proportional to the temperature at constant pressure". According to Gay Lussac's law “the pressure of a fixed mass of a gas is directly proportional to the temperature at constant volume".

#### If a sample of a gas has a volume of 500 cm^{3} at 47^{3}K, then at ____, the volume becomes 260 cm^{3}. {Assume that pressure is kept constant]

The temperature at which the volume becomes 260 cm^{3} is 245.96 K.
Using Charles' law,
V_{1}/V_{2} = T_{1}/T_{2} V_{1} = 500 cm^{3}, V_{2} = 260cm^{3}, T_{1} = 473 K T_{2} = T_{1}V_{2}/V_{1} = 473 × 260/500 = 245.96 K

#### In absolute scale, temperature is expressed in terms of _______.

In absolute scale, temperature is expressed in terms of Kelvin.

#### If air is to be a mixture of 60% nitrogen and 40% oxygen by mass, then the density of air at SATP conditions is ____.

If air is to be a mixture of 60% nitrogen and 40% oxygen by mass, then the density of air at SATP conditions is 1.193 g L^{-1}. We know that density = molar mass/molar volume Here, average molar mass of air = m_{1}x_{1 }+ m_{2}x_{2}/ x_{1} + x_{2} m_{1 }= mass of nitrogen = 28 x_{1} = percentage of nitrogen in air = 60 m_{2} = mass of oxygen = 32 x_{2} = percentage of oxygen in air = 40 Substituting, we get 28 × 60 + 32 × 40/100 = 2960/100 = 29.6 g m ol^{-1} Now, molar volume at SATP (Vm) = 24.8 L [constant] Density of air at SATP = 29.6/24.8 = 1.193 g L^{-1}

#### ____ law is defined "at constant temperature and pressure".

Avogadro's law is defined "at constant temperature and pressure". Avogadro's law states that "at constant temperature and pressure, the volume of the gas is directly proportional to its moles".

#### If 1000 mL of a gas A at 600 torr and 500 mL of gas B at 800 torr are placed in a 2L flask, then final pressure will be ____.

If 1000 mL of a gas A at 600 torr and 500 mL of gas B at 800 torr are placed in a 2L flask, then final pressure is 500 torr. PA = 1000 × 600/2000 = 300 torr PB = 500 × 800/2000 = 200 torr Ptotal = 300 + 200 = 500 torr

#### An elementary triatomic gas has vapour density 48. Its atomic mass is ____.

Its atomic mass is 32. The atomic mass of triatomic gas = molecular mass / 3 = 2 × Vapour Density / 3 = 2 × 48 / 3 = 32 u

#### If a balloon filled with an ideal gas is taken from the surface of the sea-deep to a depth of 150 m, then in terms of its original volume, its volume will be ____.

Its volume in terms of its original volume will be 6.4% of V. Pressure at the surface = 76 cm of Hg = 76 × 13.6 cm of water = 1033.6 cm = 10.3 m of water Pressure at 150 m depth = 150 + 10.3 m = 160.3 m Applying Boyle's law, P_{1}V_{1} at surface = P_{2}V_{2} at 150 m depth 10.3 × V_{1} = 160.3 × V_{2} V_{2} = 0.0642 V_{1} = 6.4% of V_{1}

#### Gases gave large inter-particle spaces between the molecules, these accounts for the high _______ of gases.

Gases gave large inter-particle spaces between the molecules, this account for the high compressibility of gases.

#### Which among the following is heavier than dry air?

Pure chlorine is heavier than dry air. The vapour density of chlorine = 35.5 The vapour density of dry air = 14.4 Therefore, pure chlorine is heavier than dry air.

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