Thermodynamic Homework Problem Check

AI Thread Summary
Mert's bedroom was 20°F, while Mort's room was calculated to be 23°F after converting -5.0°C, indicating that Mort had the colder room. In a vessel with equal masses of helium and argon, the partial pressure of helium is higher due to its lower molar mass, making statement (a) false. The average speed of helium and argon atoms is the same, validating statement (b) as true. The number of atoms differs because equal mass results in more moles of helium, making statement (c) false. Clarification was sought regarding the rms speed calculations, highlighting the difference in molecular and molar mass for the two gases.
fallen186
Messages
41
Reaction score
0
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?
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top