Question #263916

1. The gas inside a balloon has a volume of 15.0 L at a pressure of 2.0 atm Calculate the pressure of the gas if is volume is compressed to 10.0 L at the same temperature


2. A given amount of oxyçen gas has a volume of 25.0 L at a temperature of 37°C and a pressure of 10 atm Atwhat temperature would this gas occupy a volume of 22.0 L at a pressure of 10 atm?


3. The volume of a gas sample at 0 degrees C and 1.0 atm is 10.0 L How many moles of gas are contained in the sample?



4. What would be the pressure of 6.40 g oxygen gas in a vessel with a volume of 4.5 L at 20°C?



5. A sample of oxygen gas, which is saturated with water vapor, is kept in a 10-L vessel at 30°C and has a pressure of 758 Torr If the pressure of the water vapor at this temperature s 31.8 Torr, what would be the pressure of the


dry oxygen?

1
Expert's answer
2021-11-11T15:04:02-0500

1) P1 V1= P2V2

2×15= P2×10

P2= 2×1510=3atm\frac{2×15}{10}=3atm


2) T1= 37+273.15=310.15

Pressure is constant

V1T1=V2T2\frac{V_1}{T_1}= \frac{V_2}{T_2}


25310.15=22T2\frac{25}{310.15}=\frac{22}{T_2}


T2=22×310.1525=272.9KT_2 =\frac{22×310.15}{25}= 272.9K


3) At 0°C0°C and 1 atm, 1 mole of any gas occupies 22.4 litres

Number of moles of 10litre at stp

=1022.4=0.4464moles\frac{10}{22.4}= 0.4464 \> moles


4)Number of moles of oxygen

=6.4032=0.2= \frac{6.40}{32}=0.2


GasconstantR=0.082057l.atm/K.MolGas\> constant \>R=0 .082057l.atm/K.Mol

T=20+273.15= 293.15K


PV=nRT

P=0.2×0.082057×293.154.5\frac{0.2×0.082057×293.15}{4.5}

=1.069atm=1.069\>atm


5) Pressure due to Oxygen = Total pressure - vapour pressure


Pressure of dry Oxygen 75831.8=726.2torr758-31.8=726.2 torr


















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