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A given amount of oxygen gas has a volume of 25.0 L at a temperature of 37oC and a pressure of 1.0 atm. At what temperature would this gas occupy a volume of 22.0 L at a pressure of 1.0 atm?
P1V1/T1 = P2V2/T2P1 = 1 atmV1 = 25 LT1 = 37+273 = 310KP2 = 1 atmV2 = 22 LT2 = ?= 1×25/310 = 21×1/X= 0.080645 = 21/X0.080645X = 21X =...
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If 456 dm3 of krypton at 101 kPa and 21°C is compressed into a 27.0 dm3 tank at the same temperature, what is the pressure of krypton in the tank?
As at same temperature , P1 × V1 = P2 × V2Where P1 and P2 are initial and final pressures respectively and V1 and V2 are initial and final volumes...
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0.25 mol of an ideal monoatomic gas undergoes isothermal expansion from a
volume of 2.0 dm3
to 10 dm3
at 27 oC. Calculate the maximum work that can be
obtained from this process.
Wmax = nRTln(V2/V1) = 0.25 * 8.314 * (27+273) * ln(10/2) = 1004 J
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One mole of an ideal gas (group 18) at 27 ℃ and 1 atm expands adiabatically from a volume of 10 dm3 to 20 dm3. Calculate:
(i) q
(ii) w
(iii) ΔU
(iv) ΔH
(v) Final temperature of the gas.
T= 27°C= 27+273P= 1atmV1= 10dm3V2= 20dm31. q= 0, for an adiabatic expansion, heat is neither gained nor lost by the system.2. w= -P(V2-V1)w= -1(20-10)w= -10L.atm1L.atm= 101.3J-10L.atm= -10 x 101.3J= -1013J3....
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State the relevant laws which are applicable on the following observation of gas: (1) Both C3H8 and CO2 have equal rate of diffusion.(2)1 dm3 O2 and 1 dm3 H2 contains same number of moles at STP.
1) For two gases with molecular masses m1 and m2 ; and rates of diffusion R1 and R2: R1/R2 = √m2/√m1 – (1) 2) n = v/22.4
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During normal conditions the respiratory gases were mixed 25 dm3 of carbon dioxide with 10 dm3 of oxygen. Express the composition of gas mixture in the volume percentage.
Total volume: 25=10=35 dm3Volume percentage of carbon dioxide: 25/35x100=71.4%Volume percentage of oxygen: 10/35x100=28.6%
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A gas filled a 5 dm<sup>3 </sup>container at a pressure of 3.0 atm, at 301 K. If the pressure of the system is doubled, what is the new volume that the gas will occupy at 301 K?
P1 = 3.0 atmV1 = 5 dm3P2 = (3.0 * 2) = 6.0 atmV2 = unknownT1 = T2 = 301 K = const Solution:Since the temperature and amount of gas remain...
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2.0 moles of an idea has are at temperature of -133 degrees Celsius and a pressure of 2atm .what volume in dm3 will the gas occupy at that temperature? R = 0,082 atm, dm3 K -1mol-1
P = 2 atmV = unknownn = 2.0 molR = 0.082 atm dm3 K-1 mol-1T = -133°C = 140 КSolution:The Ideal Gas equation can be used: PV = nRTTo find...
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A 2.0 dm³ container contains 0.15 mol of N₂ gas and 0.20 mol of O₂ gas under a total pressure of 100 kNm⁻². Calculate the partial pressure of O₂ gas in the mixture.
Solution:Calculate the mole fraction (x) of O2 gas in the mixture:x(O2) = number moles of O2 / total number moles of gasTotal number moles of gas = Moles of...
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What is the pressure inside a cylinder when the volume changes from 14.37 dm3 to 49.31 dm3 with an initial pressure of 9.021 x 102 kPa? The temperature of the cylinder remained constant at 22.06ºC.
Answer on Question #83633, Chemistry/ General ChemistryWhat is the pressure inside a cylinder when the volume changes from 14.37 dm3 to 49.31 dm3 with an initial pressure of 9.021 x...
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