Ideal gas undergoing cycle change - find temperature

AI Thread Summary
The discussion focuses on calculating the temperature of an ideal gas at point C during a cyclic process, given that the process A-B is isothermal. The pressure and volume at point A are specified as 1.1 x 10^5 N/m^2 and 22.4 liters, respectively. The ideal gas law, PV=nRT, is referenced to relate pressure, volume, and temperature. However, without the temperature at one point, determining the temperature at point C is challenging. The conversation emphasizes the need for additional information to solve for the unknown temperature.
Amith2006
Messages
416
Reaction score
2
Sir,
One mole of an ideal gas undergoes a cycle change as shown in figure. The process A-B is isothermal. The pressure and volume at A is 1.1 x 10^5 N/m^2 and 22.4 litres respectively. What is the temperature of the gas at C?
If the temperature at one point was given the temperature at the other point could be calculated. Here only the pressure and volume are given.
 

Attachments

  • P V diagram.JPG
    P V diagram.JPG
    4.8 KB · Views: 513
Physics news on Phys.org
Amith2006 said:
Sir,
One mole of an ideal gas undergoes a cycle change as shown in figure. The process A-B is isothermal. The pressure and volume at A is 1.1 x 10^5 N/m^2 and 22.4 litres respectively. What is the temperature of the gas at C?
If the temperature at one point was given the temperature at the other point could be calculated. Here only the pressure and volume are given.
PV=nRT

AM
 
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...

Similar threads

Replies
3
Views
1K
Replies
8
Views
2K
Replies
4
Views
2K
Replies
2
Views
1K
Replies
2
Views
2K
Back
Top