A) If 4.90 L of CO2 gas at 16 ?C at 779 mmHg is used, what is the final volume,
ID: 704303 • Letter: A
Question
A) If 4.90 L of CO2 gas at 16 ?C at 779 mmHg is used, what is the final volume, in liters, of the gas at 39 ?C and a pressure of 755 mmHg , if the amount of CO2 remains the same?
B) A sample of methane (CH4) has a volume of 28 mL at a pressure of 0.74 atm . What is the final volume, in milliliters, of the gas at each of the following pressures, if there is no change in temperature and amount of gas? 0.24atm and 2.80atm
C) A gas has a volume of 3.60 L at 0 ?C. What final temperature, in degrees Celsius, is needed to change the volume of the gas to each of the following, if n and P do not change?
1.30 L , 1800mL, 61mL
Explanation / Answer
A. Initial Final
P1 = 779mmHg P2 = 755mmHg
V1 = 4.9L V2 =
T1 = 16+273 = 289K T2 = 39+273 = 312K
P1V1/T1 = P2V2/T2
V2 = P1V1T2/T1P2
= 779*4.9*312/289*755 = 5.46L
Initial Final
P1 = 0.74atm P2 = 2300Hg = 2300/760 = 3.0263atm
V1 =28ml V2 =
P1V1 = P2V2
V2 = P1V1/P2
= 0.74*28/3.0263 = 6.85ml
Initial Final
P1 = 0.74atm P2 = 81Hg = 81/760 = 0.1066atm
V1 =28ml V2 =
P1V1 = P2V2
V2 = P1V1/P2
= 0.74*28/0.1066 = 194.4ml
B. initial Final
P1 = 0.74atm P2 = 0.24atm
T1 = 28ml V2 =
P1V1 = P2V2
V2 = P1V1/P2
= 0.74*28/0.24 = 86.34ml
initial Final
P1 = 0.74atm P2 = 2.8atm
T1 = 28ml V2 =
P1V1 = P2V2
V2 = P1V1/P2
= 0.74*28/2.8 = 7.4ml
C. initial Final
V1 = 3.6L V2 = 1.3L
T1 = 0 + 273 = 273 K T2 =
V1/T1 = V2/T2
T2 = V2T1/V1
= 1.3*273/3.6 = 98.58K
= 98.58-273 = -174.420C
initial Final
V1 = 3.6L V2 = 1800ml = 1.8L
T1 = 0 + 273 = 273 K T2 =
V1/T1 = V2/T2
T2 = V2T1/V1
= 1.8*273/3.6 =136.5K
= 136.5-273 = -136.50C
nitial Final
V1 = 3.6L V2 = 61ml = 0.061L
T1 = 0 + 273 = 273 K T2 =
V1/T1 = V2/T2
T2 = V2T1/V1
= 0.061*273/3.6 =4.625K
= 4.625-273 = -268.3750C
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