Can You Check My Answers to These Gas Law Problems?

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SUMMARY

The forum discussion centers on gas law problems involving calculations of pressure, temperature, and moles of gas using the Ideal Gas Law and related equations. The first question correctly applies Boyle's Law to determine that halving the volume of a gas at constant temperature results in a pressure of 4.00 atm. The second question uses Charles's Law to find that the new temperature of a gas expanding from 50mL to 100mL at constant pressure is 546K. However, the third question's calculation of moles of oxygen gas in a 3.5L container at 0.5atm and 77°C is incorrect, as it suggests an unrealistic value of 50.225 moles, contradicting the principles of gas behavior at standard temperature and pressure (STP).

PREREQUISITES
  • Understanding of Boyle's Law and Charles's Law
  • Familiarity with the Ideal Gas Law (PV=nRT)
  • Knowledge of standard temperature and pressure (STP) conditions
  • Basic algebra for solving equations
NEXT STEPS
  • Review the Ideal Gas Law and its applications in real-world scenarios
  • Study the concepts of STP and how they relate to gas volume and moles
  • Practice solving gas law problems with varying conditions
  • Explore the implications of gas behavior under non-ideal conditions
USEFUL FOR

Chemistry students, educators, and anyone involved in gas law calculations or physical chemistry applications will benefit from this discussion.

janesmith
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Hello. If somebody has time can you check my answers?

Q1: A gas occupies a 2.0L container at 25°C and 2.0atm. If the volume of the container is halved and the temperature remains constant, what is the new pressure?

P1V1 = P2V2

P2 = 2.00 atm x (2.00L / 1.00L) = 4.00 atm

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Q2: The volume of 50mL of an ideal gas at STP increases to 100mL. If the pressure remains constant, what is the new temp?

V1/T1 = V2/T2

T2 = (100mL / 50 mL) x 273K
T2 = 546K

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Q3: How many moles of Oxygen gas, O2, are present in a 3.5L container held at a pressure of 0.5atm and a temp of 77°C?

n = PV/RT

n = (0.5atm x 3.5L) / (0.082 L•atm/mol•K x 350K)
n = 50.225 mole O2

Thanks in advanced if you take the time to check these for me.
 
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Your method is correct, I didn't check your math. You have a calculator for that!

Good luck!
 
janesmith said:
Q3: How many moles of Oxygen gas, O2, are present in a 3.5L container held at a pressure of 0.5atm and a temp of 77°C?

n = PV/RT

n = (0.5atm x 3.5L) / (0.082 L•atm/mol•K x 350K)
n = 50.225 mole O2

This MUST be wrong. At STP 1 mole of gas occupies 22.4 L. 0.5atm and 77°C is not STP, but it is also not that far - so 3.5 L should be just a fraction of a mole, not 50 moles.
 

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