Q: Question 17Taking the volume of hydrogen as calculated in Q.16 what change must be made in Kelvin (absolute) temperature to return the volume to 2500 cm3 (pressure remaining constant).

V1 = Initial volume of the gas = 50007\dfrac{5000}{7}75000 cm3T1 = Initial temperature of the gas = 273 KV2 = Final volume of the gas = 2500 cm3T2 = Final temperature of the gas = ?By Charles' Law:V1T1=V2T2\dfrac{\text{V}_1}{\text{T}_1} = \dfrac{\text{V}_2}{\text{T}_2}T1V1=T2V2Substituting the values :50007×273=2500T2T2=7×273×25005000T2=7×2732T2=3.5×273\dfrac{5000}{7 \times 273} = \dfrac{2500}{\text{T}_2} \\[1em] \text{T}_2 = \dfrac{7 \times 273 \times 2500}{5000} \\[1em] \text{T}_2 = \dfrac{7 \times 273}{2} \\[1em] \text{T}_2 = 3.5 \times 273 \\[1em]7×2735000=T22500T2=50007×273×2500T2=27×273T2=3.5×273∴ Final temperature = 3.5 times of Kelvin temperature.

Question 1

Question 1What volume will a gas occupy at 740 mm pressure which at 1480 mm occupies 500 cc ? [Temperature being constant]

Accepted Answer

V1 = Initial volume of the gas = 500 ccP1 = Initial pressure of the gas = 1480 mmP2 = Final pressure of the gas = 740 mmV2 = Final volume of the gas = ?By Boyle's Law:P1V1=P2V2\text{P}_1 \text{V}_1 = \text{P}_2 \text{V}_2P1​V1​=P2​V2​Substituting the values :1480×500=V2×740V2=1480×500740V2=1000 cc1480 \times 500 = \text{V}_2 \times 740 \[0.5em] \text{V}_2 = \dfrac{1480 \times 500}{740} \[0.5em] \text{V}_2 = 1000 \text{ cc}1480×500=V2​×740V2​=7401480×500​V2​=1000 ccTherefore, final volume of the gas = 1000 cc

Question 2

Question 4What temperature would be necessary to double the volume of a gas initially at s.t.p. if the pressure is decreased by 50% ?

Accepted Answer

Initial conditions [S.t.p.]Final conditionsP1 = Initial pressure of the gas = 1 atmP2 = Final pressure of the gas = 0.5 atmV1 = Initial volume of the gas = VV2 = Final volume of the gas = 2VT1 = Initial temperature of the gas = 273 KT2 = Final temperature of the gas = ?By Gas Law:P1×V1T1=P2×V2T2\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}T1​P1​×V1​​=T2​P2​×V2​​Substituting the values :1×V273=0.5×2VT2T2=0.5×2×273T2=273K\dfrac{1\times \text{V}}{273} = \dfrac{0.5\times\text{2\text{V}}}{\text{T}_2} \[0.5em] \text{T}_2 = 0.5\times 2 \times 273\[0.5em] \text{T}_2 = 273\text{K}2731×V​=T2​0.5×2V​T2​=0.5×2×273T2​=273KTherefore, final temperature of the gas = 273 K = 0 °C

Question 3

Question 8At a given temperature the pressure of a gas reduces to 75% of it's initial value and the volume increases by 40% of it's initial value. Find this temperature if the initial temperature was -10°C.

Accepted Answer

Initial conditions :P1 = Initial pressure of the gas = PV1 = Initial volume of the gas = VT1 = Initial temperature of the gas = -10°C = 273-10 = 263 KFinal conditions:P2 = Final pressure of the gas = reduces to 75% of P = 75100\dfrac{75}{100}10075​ of P = 34P\dfrac{3}{4}\text{P}43​PV2 = Final volume of the gas = increases by 40% of it's initial value=40100 of V + V=4V10+V=4V + 10V10=14V10= \dfrac{40}{100} \text{ of V + V} \[1em] = \dfrac{4\text{V}}{10} + \text{V} \[1em] = \dfrac{4\text{V + 10\text{V}}}{10} \[1em] = \dfrac{14\text{V}}{10}=10040​ of V + V=104V​+V=104V + 10V​=1014V​T2 = Final temperature of the gas = ?By Gas Law:P1×V1T1=P2×V2T2\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}T1​P1​×V1​​=T2​P2​×V2​​Substituting the values :P×V263=34P×14V10T2T2=34×1410×263T2=276.15K\dfrac{\text{P} \times \text{V}}{263} = \dfrac{\dfrac{3}{4}\text{P} \times\dfrac{14\text{V}}{10}}{\text{T}_2} \[0.5em] \text{T}_2 = \dfrac{3}{4} \times\dfrac{14}{10} \times 263 \[0.5em] \text{T}_2 = 276.15 \text{K}263P×V​=T2​43​P×1014V​​T2​=43​×1014​×263T2​=276.15KTherefore, final temperature of the gas = 276.15 K - 273 K = 3.15 °C

Question 4

Question 12What temperature would be necessary to double the volume of a gas, initially at S.T.P., if the pressure is decreased to 50% ?

Accepted Answer

Initial conditions [S.T.P.]Final conditionsP1 = Initial pressure of the gas = 1 atmP2 = Final pressure of the gas = 0.5 atmV1 = Initial volume of the gas = VV2 = Final volume of the gas = 2VT1 = Initial temperature of the gas = 273 KT2 = Final temperature of the gas = ?By Gas Law:P1×V1T1=P2×V2T2\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}T1​P1​×V1​​=T2​P2​×V2​​Substituting the values :1×V273=0.5×2VT2T2=0.5×2×273T2=273 K\dfrac{1\times \text{V}}{273} = \dfrac{0.5\times\text{2\text{V}}}{\text{T}_2} \[0.5em] \text{T}_2 = 0.5\times 2 \times 273\[0.5em] \text{T}_2 = 273\text{ K}2731×V​=T2​0.5×2V​T2​=0.5×2×273T2​=273 K∴ Final temperature of the gas = 273 K

Question 5

Question 13At 0°C and 760 mm Hg pressure, a gas occupies volume of 100 cm3. Kelvin temperature of the gas is increased by one-fifth and the pressure is increased one and a half times. Calculate the final volume of the gas.

Accepted Answer

Initial conditions [S.T.P.] :P1 = Initial pressure of the gas = 760 mm HgV1 = Initial volume of the gas = 100 cm3T1 = Initial temperature of the gas = 0°C = 273 KFinal conditions :P2 (Final pressure) = increased one and a half times of P1= (1 + 12\dfrac{1}{2}21​) of 760= 32\dfrac{3}{2}23​ x 760= 3 x 380= 1140 mm HgV2 (Final volume) = ?T2 (Final temperature) = increased by one-fifth of 273 K= (1+15)\Big(1 + \dfrac{1}{5}\Big)(1+51​) of 273= 65\dfrac{6}{5}56​ x 273 = 16385\dfrac{1638}{5}51638​By Gas Law:P1×V1T1=P2×V2T2\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}T1​P1​×V1​​=T2​P2​×V2​​Substituting the values :760×100273=1140×V216385760×100273=5×1140×V21638V2=760×100×1638273×5×1140V2=80 cm3\dfrac{760 \times 100 }{273} = \dfrac{1140 \times \text{V}_2}{\dfrac{1638}{5}} \[1em] \dfrac{760 \times 100 }{273} = \dfrac{5 \times 1140 \times \text{V}_2}{1638} \[1em] \text{V}_2 = \dfrac{760 \times 100 \times 1638}{273 \times 5 \times 1140} \[1em] \text{V}_2 = 80 \text{ cm}^3273760×100​=51638​1140×V2​​273760×100​=16385×1140×V2​​V2​=273×5×1140760×100×1638​V2​=80 cm3∴ Final volume of the gas = 80 cm3

Question 6

Question 14It is found, on heating a gas, its volume increases by 50% and pressure decreases to 60% of its original value. If the original temperature was -15°C, find the temperature to which it was heated?

Accepted Answer

Initial conditions [S.T.P.] :P1 = Initial pressure of the gas = PV1 = Initial volume of the gas = VT1 = Initial temperature of the gas = -15°C = -15 + 273 = 258 KFinal conditions :P2 (Final pressure) = pressure decreases to 60% of its original value= 60100\dfrac{60}{100}10060​ of P= 35\dfrac{3}{5}53​PV2 (Final volume) = volume increases by 50% of its original value= 1 + 50100\dfrac{50}{100}10050​ of V= 100+50100\dfrac{100 + 50}{100}100100+50​ of V= 150100\dfrac{150}{100}100150​ of V= 32\dfrac{3}{2}23​VT2 (Final temperature) = ?By Gas Law:P1×V1T1=P2×V2T2\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}T1​P1​×V1​​=T2​P2​×V2​​Substituting the values :PV 258=35P×32VT2T2=3×3×2585×2=232210T2=232.2K\dfrac{\text{PV }}{258} = \dfrac{\dfrac{3}{5} \text{P}\times\dfrac{3}{2}\text{V}}{\text{T}_2} \[1em] \text{T}_2 = \dfrac{3\times 3\times 258}{5 \times 2} = \dfrac{2322}{10} \[1em] \text{T}_2 = 232.2 \text{K} \[1em]258PV ​=T2​53​P×23​V​T2​=5×23×3×258​=102322​T2​=232.2K∴ Final temperature of the gas = 232.2 - 273 = -40.8°C

Question 7

Question 15A certain mass of a gas occupies 2 Litres at 27°C and 100 pascal. Find the temperature when volume and pressure become half of their initial values.

Accepted Answer

Initial conditions [S.T.P.] :P1 = Initial pressure of the gas = 100 PascalV1 = Initial volume of the gas = 2 LitresT1 = Initial temperature of the gas = 27°C = 27 + 273 = 300 KFinal conditions :P2 (Final pressure) = 50 PascalV2 (Final volume) = 1 litreT2 (Final temperature) = ?By Gas Law:P1×V1T1=P2×V2T2\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}T1​P1​×V1​​=T2​P2​×V2​​Substituting the values :100×2300=50×1T2T2=50×32T2=75K\dfrac{100 \times 2}{300} = \dfrac{50 \times 1}{\text{T}_2} \[1em] \text{T}_2 = \dfrac{50\times 3}{2} \[1em] \text{T}_2 = 75 \text{K} \[1em]300100×2​=T2​50×1​T2​=250×3​T2​=75K∴ Final temperature of the gas = 75 - 273 = -198°C

Question 8

Question 162500 cm3 of hydrogen is taken at S.T.P. The pressure of this gas is further increased by two and and a half times (temperature remaining constant). What volume will hydrogen occupy now?

Accepted Answer

V1 = 2500 cm3P1 = 760 mm HgP2 = increased by two and a half times= (1 + 2122\dfrac{1}{2}221​ ) times of 760 mm Hg= (1 + 52\dfrac{5}{2}25​ ) times of 760 mm Hg= ( 72\dfrac{7}{2}27​ x 760) mm HgV2 = ?By Boyle's Law:P1V1=P2V2\text{P}_1 \text{V}_1 = \text{P}_2 \text{V}_2P1​V1​=P2​V2​Substituting the values :760×2500=72×760×V2V2=50007760 \times 2500 = \dfrac{7}{2} \times 760 \times \text{V}_2 \[1em] \text{V}_2 = \dfrac{5000}{7} \[0.5em]760×2500=27​×760×V2​V2​=75000​∴ Final volume = 50007\dfrac{5000}{7}75000​ cm3

Question 9

Question 17Taking the volume of hydrogen as calculated in Q.16 what change must be made in Kelvin (absolute) temperature to return the volume to 2500 cm3 (pressure remaining constant).

Accepted Answer

V1 = Initial volume of the gas = 50007\dfrac{5000}{7}75000​ cm3T1 = Initial temperature of the gas = 273 KV2 = Final volume of the gas = 2500 cm3T2 = Final temperature of the gas = ?By Charles' Law:V1T1=V2T2\dfrac{\text{V}_1}{\text{T}_1} = \dfrac{\text{V}_2}{\text{T}_2}T1​V1​​=T2​V2​​Substituting the values :50007×273=2500T2T2=7×273×25005000T2=7×2732T2=3.5×273\dfrac{5000}{7 \times 273} = \dfrac{2500}{\text{T}_2} \[1em] \text{T}_2 = \dfrac{7 \times 273 \times 2500}{5000} \[1em] \text{T}_2 = \dfrac{7 \times 273}{2} \[1em] \text{T}_2 = 3.5 \times 273 \[1em]7×2735000​=T2​2500​T2​=50007×273×2500​T2​=27×273​T2​=3.5×273∴ Final temperature = 3.5 times of Kelvin temperature.

Question 10

Question 18A given amount of gas A is confined in a chamber of constant volume. When the chamber is immersed in a bath of melting ice, the pressure of the gas is 100 cm Hg.(a)	What is the temperature, when the pressure is 10 cm Hg?(b) What will be the pressure, when the chamber is brought to 100°C?

Accepted Answer

(a) P1 = Initial pressure of the gas = 100 cm HgT1 = Initial temperature of the gas = 273 KP2 = Final pressure of the gas = 10 cm HgT2 = Final temperature of the gas = ?V1 = V2 = V [constant volume]By Gas Law:P1×V1T1=P2×V2T2\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}T1​P1​×V1​​=T2​P2​×V2​​Substituting the values :100×V273=10×VT2T2=27310T2=27.3K\dfrac{100 \times \text{V}}{273} = \dfrac{10 \times \text{V}}{\text{T}_2} \[1em] \text{T}_2 = \dfrac{ 273}{10} \[1em] \text{T}_2 = 27.3 \text {K} \[1em]273100×V​=T2​10×V​T2​=10273​T2​=27.3K∴ When the pressure is 10 cm Hg, the temperature is 27.3 K.(b) P1 = Initial pressure of the gas = 100 cm HgT1 = Initial temperature of the gas (ice point) = 273 KP2 = Final pressure of the gas = ?T2 = Final temperature of the gas = 100°C = 273 + 100 = 373 KV1 = V2 = V [constant volume]By Gas Law:P1×V1T1=P2×V2T2\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}T1​P1​×V1​​=T2​P2​×V2​​Substituting the values :100×V273=P2×V373P2=37300273P2=136.63 cm of Hg\dfrac{100 \times \text{V}}{273} = \dfrac{\text{P}_2 \times \text{V}}{373} \[1em] \text{P}_2 = \dfrac{ 37300}{273} \[1em] \text{P}_2 = 136.63 \text { cm of Hg} \[1em]273100×V​=373P2​×V​P2​=27337300​P2​=136.63 cm of Hg∴ When the chamber is brought to 100°C, the pressure is 136.63 cm Hg.

Question 11

Question 19A gas is to be filled from a tank of capacity 10,000 litres into cylinders each having capacity of 10 litres. The condition of the gas in the tank is as follows:(a) pressure inside the tank is 800 mm of Hg.(b)	temperature inside the tank is -3°C.When the cylinder is filled, the pressure gauge reads 400 mm of Hg and the temperature is 270 K. Find the number of cylinders required to fill the gas.

Accepted Answer

Initial conditions :P1 = Initial pressure of the gas = 800 mm of HgV1 = Initial volume of the gas = 10,000 litresT1 = Initial temperature of the gas = -3°C = -3 + 273 = 270 KFinal conditions:P2 (Final pressure) = 400 mm of HgV2 (Final volume) = 10 litres x n cylindersT2 (Final temperature) = 270 Kwhere n is the number of cylinders = ?By Gas Law:P1×V1T1=P2×V2T2\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}T1​P1​×V1​​=T2​P2​×V2​​Substituting the values :800×10,000270=400×10×n270n=800×10,000×270400×10×270n=800×104n=2000\dfrac{800 \times 10,000}{270} = \dfrac{400 \times 10 \times n}{270} \[1em] \text{n} = \dfrac{800 \times 10,000 \times 270}{400 \times 10 \times 270} \[1em] \text{n} = \dfrac{800\times 10}{4} \[1em] \text{n} = 2000 \[1em]270800×10,000​=270400×10×n​n=400×10×270800×10,000×270​n=4800×10​n=2000∴ Number of cylinders = 2000.

Question 12

Question 21Calculate the temperature at which the volume of a gas is doubled if its pressure at the same time increases from 70 cm of Hg to 80 cm of Hg.

Accepted Answer

Initial conditions :P1 = Initial pressure of the gas = 70 cm of HgV1 = Initial volume of the gas = VT1 = Initial temperature of the gas = 273 KFinal conditions:P2 (Final pressure) = 80 cm of HgV2 (Final volume) = 2VT2 (Final temperature) = ?By Gas Law:P1×V1T1=P2×V2T2\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}T1​P1​×V1​​=T2​P2​×V2​​Substituting the values :70×V273=80×2VT2T2=80×2V×27370×VT2=16×2737T2=624 K\dfrac{70 \times \text{V}}{273} = \dfrac{80 \times \text{2V}}{\text{T}_2} \[1em] \text{T}_2 = \dfrac{80 \times \text{2V} \times 273}{70 \times \text{V}} \[1em] \text{T}_2 = \dfrac{16 \times 273}{7} \[1em] \text{T}_2 = 624 \text{ K} \[1em]27370×V​=T2​80×2V​T2​=70×V80×2V×273​T2​=716×273​T2​=624 K∴ Final temperature of the gas = 624 - 273 = 351°C

Question 13

Question 22A certain gas at -15°C is heated. Its volume increases by 50% and pressure decreases to 75% of its original reading. Calculate the temperature to which it was heated.

Accepted Answer

Initial conditions [S.T.P.] :P1 = Initial pressure of the gas = PV1 = Initial volume of the gas = VT1 = Initial temperature of the gas = -15°C = -15 + 273 = 258 KFinal conditions :P2 (Final pressure) = pressure decreases to 75% of its original value=75100 of P=34P=0.75P= \dfrac{75 }{100} \text{ of P} \[1em] = \dfrac{3 }{4} \text{P} \[1em] = 0.75 \text {P}=10075​ of P=43​P=0.75PV2 (Final volume) = Volume increases by 50% of its original value=1+50100 of V=100+50100 of V=150100of V=32V=1.5V= \text {1} + \dfrac{50}{100} \text{ of V} \[1em] = \dfrac{100 + 50}{100} \text{ of V} \[1em] = \dfrac{150}{100} \text{of V} \[1em] = \dfrac{3}{2} \text{V} = 1.5 \text{V}=1+10050​ of V=100100+50​ of V=100150​of V=23​V=1.5VT2 (Final temperature) = ?By Gas Law:P1×V1T1=P2×V2T2\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}T1​P1​×V1​​=T2​P2​×V2​​Substituting the values :PV258=0.75P×1.5VT2\dfrac{\text{P}\text{V}}{258} = \dfrac{0.75\text{P}\times\text{1.5V}}{\text{T}_2}258PV​=T2​0.75P×1.5V​Solve for T2:T2=1.125×258T2=290.25K\text {T}_2 = 1.125 \times 258 \[1em] \text {T}_2 = 290.25 \text {K}T2​=1.125×258T2​=290.25K∴ Final temperature of the gas = 290.25 - 273 = 17.25°C

Question 14

Question 23Calculate the volume occupied by 2 g of hydrogen at 27°C and 4 atmosphere pressure, if at S.T.P. it occupies 22.4 litres.

Accepted Answer

Initial conditions [S.T.P.] :P1 = Initial pressure of the gas = 1 atmV1 = Initial volume of the gas = 22.4 lT1 = Initial temperature of the gas = 273 KFinal conditions :P2 (Final pressure) = 4 atmT2 (Final temperature) = 27°C = 27 + 273 = 300 KV2 (Final volume) = ?By Gas Law:P1×V1T1=P2×V2T2\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}T1​P1​×V1​​=T2​P2​×V2​​Substituting the values :1×22.4273=4×V2300V2=300×22.4273×4V2=6.15 l\dfrac{1 \times 22.4}{273} = \dfrac{4 \times \text{V}_2}{300} \[1em] \text{V}_2 = \dfrac{300 \times 22.4}{273 \times 4} \[1em] \text{V}_2 = 6.15 \text{ l} \[1em]2731×22.4​=3004×V2​​V2​=273×4300×22.4​V2​=6.15 l∴ Volume occupied = 6.15 l

Question 15

Question 24Which will have greater volume when the following gases are compared at S.T.P.(a)	1.2 lit N2 at 25°C and 748 mm Hg(b) 1.25 lit O2 at S.T.P.?

Accepted Answer

1.2 lit N2 at 25°C and 748 mm HgInitial conditions :P1 = Initial pressure of the gas = 748 mm HgV1 = Initial volume of the gas = 1.2 litT1 = Initial temperature of the gas = 25°C = 25 + 273 = 298 KFinal conditions :P2 (Final pressure) = 760 mm of HgT2 (Final temperature) = 273 KV2 (Final volume) = ?By Gas Law:P1×V1T1=P2×V2T2\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}T1​P1​×V1​​=T2​P2​×V2​​Substituting the values :748×1.2298=760×V2273V2=748×1.2×273298×760V2=1.081 lit\dfrac{748 \times 1.2}{298} = \dfrac{760 \times \text{V}_2}{273} \[1em] \text{V}_2 = \dfrac{748 \times 1.2\times 273}{298 \times 760} \[1em] \text{V}_2 = 1.081 \space \text{lit} \[1em]298748×1.2​=273760×V2​​V2​=298×760748×1.2×273​V2​=1.081 lit∴ Volume of N2 at S.T.P. = 1.081 lit and volume of O2 at S.T.P. is 1.25 lit. Hence, volume of O2 at S.T.P. is greater than N2 at S.T.P.

Question 16

Question 2550 cm3 of hydrogen is collected over water at 17°C and 750 mm Hg pressure. Calculate the volume of dry gas at S.T.P. The water vapour pressure at 17°C is 14 mm Hg.

Accepted Answer

Initial conditions :P1 = 750 - water vapour pressure = 750 - 14 = 736 mm HgV1 = Initial volume of the gas = 50 cm3T1 = Initial temperature of the gas = 17°C = 17 + 273 = 290 KFinal conditions :P2 (Final pressure) = 760 mm of HgT2 (Final temperature) = 273 KV2 (Final volume) = ?By Gas Law:P1×V1T1=P2×V2T2\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}T1​P1​×V1​​=T2​P2​×V2​​Substituting the values :736×50290=760×V2273V2=736×50×273290×760V2=45.58 cm3\dfrac{736 \times 50}{290} = \dfrac{760 \times \text{V}_2}{273} \[1em] \text{V}_2 = \dfrac{736 \times 50 \times 273}{290 \times 760} \[1em] \text{V}_2 = 45.58 \text{ cm}^3 \[1em]290736×50​=273760×V2​​V2​=290×760736×50×273​V2​=45.58 cm3∴ Volume of dry gas = 45.6 cm3

Question 17

Question 26(a) Calculate the volume of S.T.P. of 380 cm3 of carbon dioxide at 91°C and 72 mm of Hg pressure.(b) What will be its volume at 273 K and 70 cm of Hg pressure ?

Accepted Answer

(a)Initial conditionsFinal conditionsP1 = Initial pressure of the gas = 72 mm HgP2 = Final pressure of the gas = 760 mm HgV1 = Initial volume of the gas = 380 cm3V2 = Final volume of the gas = ?T1 = Initial temperature of the gas = 91°C = 91 + 273 = 364 KT2 = Final temperature of the gas = 273 KBy Gas Law:P1×V1T1=P2×V2T2\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}T1​P1​×V1​​=T2​P2​×V2​​Substituting the values :72×380364=760×V2273V2=72×380×273364×760V2=27 cm3\dfrac{72 \times 380}{364} = \dfrac{760 \times \text{V}_2}{273} \[1em] \text{V}_2 = \dfrac{72 \times 380 \times 273}{364\times 760} \[1em] \text{V}_2 = 27 \text{ cm}^3 \[1em]36472×380​=273760×V2​​V2​=364×76072×380×273​V2​=27 cm3(b)Initial conditionsFinal conditionsP1 = Initial pressure of the gas = 72 mm of HgP2 = Final pressure of the gas = 70 cm Hg = 700 mm of HgV1 = Initial volume of the gas = 380 cm3V2 = Final volume of the gas = ?T1 = Initial temperature of the gas = 91°C = 91 + 273 = 364 KT2 = Final temperature of the gas = 273 KBy Gas Law:P1×V1T1=P2×V2T2\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}T1​P1​×V1​​=T2​P2​×V2​​Substituting the values :72×380364=700×V2273V2=72×380×273364×700V2=29.314 cm3\dfrac{72 \times 380}{364} = \dfrac{700 \times \text{V}_2}{273} \[1em] \text{V}_2 = \dfrac{72 \times 380 \times 273}{364\times 700} \[1em] \text{V}_2 = 29.314 \text{ cm}^3 \[1em]36472×380​=273700×V2​​V2​=364×70072×380×273​V2​=29.314 cm3

Question 18

Question 27Calculate the volume of dry air at S.T.P. that occupies 28 cm3 at 14°C and 750 mm Hg pressure when saturated with water vapour. The vapour pressure of water at 14°C is 12 mm Hg.

Accepted Answer

Initial conditions :P1 = 750 - water vapour pressure = 750 - 12 = 738 mm HgV1 = Initial volume of the gas = 28 cm3T1 = Initial temperature of the gas = 14°C + 273 = 287 KFinal conditions [S.T.P.] :P2 (Final pressure) = 760 mm of HgT2 (Final temperature) = 273 KV2 (Final volume) = ?By Gas Law:P1×V1T1=P2×V2T2\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}T1​P1​×V1​​=T2​P2​×V2​​Substituting the values :738×28287=760×V2273V2=738×28×273287×760V2=25.86 cm3\dfrac{738 \times 28 }{287} = \dfrac{760 \times \text{V}_2}{273} \[1em] \text{V}_2 = \dfrac{738 \times 28 \times 273}{287 \times 760} \[1em] \text{V}_2 = 25.86 \text{ cm}^3 \[1em]287738×28​=273760×V2​​V2​=287×760738×28×273​V2​=25.86 cm3∴ Volume of dry gas = 25.9 cm3

Question 19

Question 28L.P.G. cylinder can withstand a pressure of 14.9 atmosphere. The pressure gauge of the cylinder indicates 12 atmosphere at 27°C. Due to a sudden fire in the building the temperature rises. At what temperature will the cylinder explode?

Accepted Answer

P1 = Initial pressure of the gas = 12 atmT1 = Initial temperature of the gas = 27°C = 27 + 273 = 300 KP2 = Final pressure of the gas = 14.9 atmT2 = Final temperature of the gas = ?V1 = V2 = V [constant volume]By Gas Law:P1×V1T1=P2×V2T2\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}T1​P1​×V1​​=T2​P2​×V2​​Substituting the values :12×V300=14.9×VT2T2=14.9×30012T2=372.5K\dfrac{12\times\text{V}}{300} = \dfrac{14.9\times\text{V}}{\text{T}_2} \[1em] \text{T}_2 = \dfrac{14.9 \times 300}{12}\[1em] \text{T}_2 = 372.5 \text {K} \[1em]30012×V​=T2​14.9×V​T2​=1214.9×300​T2​=372.5K∴ The temperature at which the cylinder will explode = 99.5°C.

Question 20

Question 2922.4 litres of a gas weighs 70 g at S.T.P. Calculate the weight of the gas if it occupies a volume of 20 litres at 27°C and 700 mm Hg of pressure.

Accepted Answer

Initial conditions :P1 = 700 mm HgV1 = Initial volume of the gas = 20 litresT1 = Initial temperature of the gas = 27°C + 273 = 300 KFinal conditions [S.T.P.] :P2 (Final pressure) = 760 mm of HgT2 (Final temperature) = 273 KV2 (Final volume) = ?By Gas Law:P1×V1T1=P2×V2T2\dfrac{\text{P}_1\times\text{V}_1}{\text{T}_1} = \dfrac{\text{P}_2\times\text{V}_2}{\text{T}_2}T1​P1​×V1​​=T2​P2​×V2​​Substituting the values :700×20300=760×V2273V2=7×2×2733×76V2=16.76 l\dfrac{700 \times 20 }{300} = \dfrac{760 \times \text{V}_2}{273} \[1em] \text{V}_2 = \dfrac{7 \times 2 \times 273}{3\times 76} \[1em] \text{V}_2 = 16.76 \space l \[1em]300700×20​=273760×V2​​V2​=3×767×2×273​V2​=16.76 l22.4 l of gas at S.T.P. weighs 70 g∴ 16.76 l of gas at S.T.P. will weigh= 70×16.7622.4\dfrac{70 \times 16.76}{22.4}22.470×16.76​ = 52.38 g∴ Weight of gas = 52.38 g

Study of Gas Laws

Additional Questions

Exercise 7 Numericals

Exercise 7 Structuredapplicationskill Type

Numericals Based On Boyles Law

Numericals Based On Charles Law

Questions

Unit Test Paper 7 — Study of Gas Laws