Change in Vrms, given initial P and V, find energy added.

AI Thread Summary
The discussion revolves around calculating the energy transferred to an ideal diatomic gas during a process where pressure is proportional to volume. The initial conditions include a volume of 2.14 m³ and a pressure of 1 atm, with the gas's rms speed doubling by the end of the process. Participants express confusion over the relationship between pressure and volume, particularly regarding PV diagrams and isotherms. Key equations mentioned include the rms speed formula and the heat transfer equation, but the application of these equations remains unclear for some users. Ultimately, the discussion highlights the challenges in understanding the implications of the pressure-volume relationship in this context.
MattRob
Messages
208
Reaction score
29

Homework Statement


One mole of an ideal diatomic gas with C_{v} = \frac{5R}{2} occupies a volume Vi = 2.14 m^{3} at a pressure P_{i}=1 atm. The gas undergoes a process in which the pressure is proportional to the volume. At the end of the process, it is found that the rms speed of the gas molecules has doubled from its initial value. Determine the amount of energy transferred to the gas by heat.

Homework Equations


? (If I knew this, I wouldn't be here)

The Attempt at a Solution


With some searching, I found the equations V_{rms} = \sqrt{\frac{3RT}{M}} and Q = MC_{v}\Delta T

Where Q is a change in energy, V_{rms} is the average rms velocity, T is temperature, M is Molar Mass, and \Delta T is change in temperature.

With some substitution, we eventually found that we can get;
Q = M^{2}V_{rms_{initial}}^{2} \frac{5}{2}

Which... Doesn't help at all. So, yeah.

The question mentions that the pressure is proportional to the volume, but I'm still a bit confused about PV diagrams, and why everything doesn't stay on Isotherms, so I'm really not sure what it means when it says that pressure is proportional to volume. Does it mean kP = V or \frac{P}{V} = 1, or what? So... Yeah.
 
Physics news on Phys.org
Problem solved. Got some unexpected help.
 
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