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Under a pressure of 200 kPa, a confined gas has a volume of 2,500 cubic meters. The pressure acting on the gas is increased to 500 kPa. What is the new volume of the gas if the temperature remains the same?
Use the : $P_1 V_1 = P_2 V_2$ . substituting and making the units consistent: $(200xx10^3)(2500)=(500xx10^3)(V_2)$. solve for $V_2$ .
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Ethene undergoes incomplete combustion to form carbon dioxide, carbon monoxide and water vapour. If the ratio of carbon dioxide to carbon monoxide formed is 9:1 when x cm^3 of ethene is burnt, what is the volume of oxygen gas consumed in the reaction?
Ethene undergoes incomplete combustion to form carbon dioxide, carbon monoxide and water vapour. The balanced equation of this incomplete combustion reaction is as follows. $C_2H_4(g)+5/2O_2(g)->CO_2(g)+CO(g)+2H_2O(g)$ But as per question the...
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A breathing mixture used by deep-sea divers contains helium, oxygen, and carbon dioxide What is the partial pressure of oxygen at $101.4$ $kPa$ if $P_(He)$ = $82.5$ $kPa$ and $P_(CO2)$ = $.4$ $kPa$?
The total pressure is the sum of the partial pressures of the given gases. $P_"Total"=P_"He"+P_"CO2"+P_"O2"$ Rearrange the equation to isolate $P_"O2"$, substitute the given values into the equation and solve....
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A container of gas with a pressure of 450 kPa contains three different gasses -hydrogen, oxygen, and nitrogen. If the partial pressure of the hydrogen is 210 kPa and the partial pressure of the oxygen is 125 kPa, what is the partial pressure of nitrogen?
According to Dalton's law of partial pressures, the total pressure is equivalent to the sum of the pressures of the individual gases (if they were present alone). Therefore, the...
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The pressure of gases inside a rigid container is $"99000 kPa"$. The composition of the gases is $68%$ nitrogen gas, $23%$ oxygen gas, and $9%$ neon gas. What is the partial pressure of each gas?
Multiply the percentage in decimal form of each gas by $"99000 kPa"$. Nitrogen $0.68xx99000color(white)(.)"kPa"$$=$$"67320 kPa"$ Oxygen $0.23xx99000color(white)(.)"kPa"$$=$$"22770 kPa"$ Neon $0.09xx99000color(white)(.)"kPa"$$=$$"8910 kPa"$ The sum of the partial pressures equals the total...
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A sample of carbon dioxide gas at a pressure of 1.07 atm and a temperature of 166 °C, occupies a volume of 686 mL. If the gas is heated at constant pressure until its volume is 913 mL, the temperature of the gas sample will be ? °C.
The idea here is that the volume and the temperature of a gas have a direct relationship when the pressure and the number of moles of gas are being kept...
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A sample of gas has a volume of 25 L at a pressure of 200 kPa and a temperature of 25°C. What would be the volume if the pressure were increased to 250 kPa and the temperature were decreased to 0°C?
$(P_1V_1)/T_1=(P_2V_2)/T_2$ You need to convert degC to Kelvin by adding 273: $:.(200xx25)/298=(250xxV_2)/273$ $:.V_2=(200xx25xx273)/(298xx250)$ $:.V_2=18.32"L"$
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The pressure of 250 kPa acting on 15 cubic meters of gas at a temperature of 100 K is increased to 500 kPa and the volume is increased to 30 cubic meters. What is the temperature of the gas at this new pressure and volume?
This is an example of the . The equation to use is $(P_1V_1)/(T_1)=(P_2V_2)/(T_2)$. Given Initial pressure, $P_1="250 kPa"$ Initial volume, $V_1="15 m"^3"$ Initial temperature, $T_1="100 K"$ Final pressure, $P_2="500 kPa"$...
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If a tire has a pressure of 11.9 Pa, a volume of 2.8 L at a temperature of 319 K, and the pressure increases to 13.4 kPa when the temperature rises to 325 K, what is the new volume of the tire?
We use the , $(P_1V_1)/T_1=(P_2V_2)/T_2$ Thus $V_2=(P_1*T_2*V_1)/(P_2*T_1)$ $=$ $(11.9*Paxx325*Kxx2.8*L)/(13.4*Paxx319*K)$ So pressure seems to win over temperature here inasmuch as the starting volume has been reduced.
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The solubility of carbon dioxide in water is 0.161 g $CO_2$ in 100 mL of water at 20°C and 1.00 atm. A soft drink is carbonated with carbon dioxide gas at 5.50 atm pressure. What is the solubility of carbon dioxide in water at this pressure?
$"8.855 g/L"$ This is just asking you to calculate the new molar (i.e. molar solubility) at a different pressure, i.e. use the . (We'll end up converting...
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