1. A gas was compressed unde conditions. The original gas pressure r constant te
ID: 556508 • Letter: 1
Question
1. A gas was compressed unde conditions. The original gas pressure r constant temperature was 742 mmHg and the expansion was accomplished by changing the volume from 6.50 L to 4.00L. What is the pressure (mmHg ) of the compressed gas? A. 0.0400 mmHg B. 471 mmHg C. 1 mmHg D. 1206 mmHg 2. A sample ofoxygen gas (02) has a volume of 22.41 L, containing 1 .00 mols of oxygen gas (O.). If some oxygen is removed from the sample, changing its volume to 10.00 L how many moles of gas are now in the container? A. 1.27 mols B. 0.446 mols C. 4.92 mols D. 6.45 mols 3. How many mols ofoxygen (02) were removed from the sample in question #2? A. 3.8 mols B. 1.80 mols C. 0.554 mols D. 3.2 mols -. The volume of a gas was changed from 2.50 L at 25.0 °C to a new volume, using a U- temperature of 343 K. A. 3.75 L What was the new volume of the gas? B. 2.32L C. 6.52 L D. 2.88 L 5. Which law is used in question #1? A. Avogadro's B. Boyle's C. Charles's D. Dalton's 6. Which law is used in question #2? A. Avogadro's B. Boyle's C. Charles's D. Dalton'sExplanation / Answer
2) Initially the container has 1.00mol of O2 gas at a volume of 22.41 L
now some gas is expelled as the volume is changed to 10L
Assuming pressure remaining constant(open vessel),
V1/n1 = V2/n2 (Avogadro's law)
22.41L/ 1.00mol = 10L/ n2
Thus n2 = th emoles remaining = 0.446 mol
OPtion B
3) The moles of O2 escaped = inital moles - moles of gas remained
= 1.00 - 0.446
= 0.5535 mol
OPTION C
4) At constant pressure V1/T1 = V2/T2 (Charles' law)
2.5L/ 298K = V2 /343K
Thus volume of gas at 343K = 2.877L
OPTION D
5) The law relating P and V at constant temperature is BOYLE'S LAW
OPTION B
6) the law that relates number of moles of gas and volume is AVOGADRO'S LAW
OPTION A
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