Thermodynamic Homework Problem Check

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SUMMARY

The discussion focuses on two thermodynamic homework problems involving temperature conversions and gas behavior. In the first problem, Mert's room at 20°F is warmer than Mort's room at -5.0°C, which converts to 23°F, confirming that Mort has the colder room. The second problem examines the behavior of helium and argon in a vessel, concluding that while helium exerts a higher partial pressure due to its lower molar mass, both gases have the same root mean square (rms) speed when considering their respective molecular masses.

PREREQUISITES
  • Understanding of temperature conversion formulas, specifically t_{F} = \frac{9}{5}*t_{c} + 32°F
  • Knowledge of the ideal gas law, PV=nRT
  • Familiarity with concepts of partial pressure and molar mass
  • Basic understanding of root mean square speed calculations, v_{rms} = \sqrt{\frac{3kT}{m}} and v_{rms} = \sqrt{\frac{3RT}{M}}
NEXT STEPS
  • Research the implications of molar mass on gas behavior and partial pressures
  • Study the derivation and applications of the ideal gas law in various conditions
  • Explore the differences between molecular mass and molar mass in thermodynamic calculations
  • Learn more about the kinetic theory of gases and its relation to temperature and pressure
USEFUL FOR

Students studying thermodynamics, physics enthusiasts, and educators looking to clarify concepts related to gas behavior and temperature conversions.

fallen186
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Can you check over my answers to see if my reasoning is correct?

Homework Statement


#3) "Yesterday I woke up and it was 20°F in my bedroom," said Mert to his old friend Mort. "That's nothing," replied Mort. "My room was -5.0°C." Who had the colder room, Mert or Mort?


Homework Equations


t_{F} = \frac{9}{5}*t_{c} + 32°F


The Attempt at a Solution


Mort's -5.0°C Room conversion:
t_{F} = \frac{9}{5}*t_{c} + 32°F

t_{F} = \frac{9}{5}*(-5.0C) + 32°F

t_{F} = -9C + 32°F

t_{F} = 23°F

Mort = 23°F
Mert = 20°F

Mert has the colder room.

Homework Statement


#7) If a vessel contains equal amounts, by mass, of helium and argon, which of the following are true?
(a) The partial pressure exerted by each of the two gases on the walls of the container is the same
(b) The average speed of a helium atom is the same as that of an argon atom.
(c) The number of helium atoms and argon atoms in the vessel are equal.
(d) None of the above.


Homework Equations


(a) PV=nRT


The Attempt at a Solution


(a) False PV = nRT since V, R, and T are the same then what really matters is the relationship between P and n. If n is greater than P increases as well (direct relationship). Since both gases are the same by mass that means there is more moles of helium due to its small molar mass. Helium thus has more partial pressure than argon.
(b) True The equation v_{rms} = \sqrt_{\frac{3kT}{m}}. Since helium and argon have the same mass their speeds are the same.
(c) False . since they're are equal by mass and have different molar masses they have different amount of moles and different amount of atoms.
(d) False. It's false because b is true

Question about the rms speed of molecules.

Book explanation:
The rms speed of a molecule of a gas is related to the absolute temperature by v_{rms} = \sqrt_{(v^{2})_{av}} = \sqrt_{\frac{3kT}{m}} = \sqrt_{\frac{3RT}{M}} where m is the mass of the molecule and M is the molar mass.

For part b of the previous question. If i used
\sqrt_{\frac{3kT}{m}} then I get the same speed but if i use

\sqrt_{\frac{3RT}{M}} then i get a different speed.

Can someone help me explain this?
 
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Surely both m and M are different for Helium and Argon.
Helium has a molar mass of 4 g/mol and a molecular mass of 4*1.67*10^-27 kg.
Argon has a molar mass of 40 g/mol and a molecular mass of 40*1.67*10^-27 kg
 
Delphi51 said:
Surely both m and M are different for Helium and Argon.
Helium has a molar mass of 4 g/mol and a molecular mass of 4*1.67*10^-27 kg.
Argon has a molar mass of 40 g/mol and a molecular mass of 40*1.67*10^-27 kg

Thanks. This helps a lot.
 

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