What Is the Work Done by the Gas and the Final Gas Temperature?

AI Thread Summary
The discussion revolves around calculating the work done by an ideal gas in a piston-cylinder device during a heating process. The work done by the gas is determined to be -132.54 J, indicating that the gas does work on the piston, which is consistent with the negative sign. The final gas temperature is calculated to be 450 K, with a total heat transfer of 464.26 J during the process. The use of the specific heat at constant volume (718 J/kg·K) is justified as the problem focuses on changes in internal energy rather than enthalpy. The constant pressure condition is clarified, emphasizing that the piston can move freely, allowing for pressure stability.
Legendon
Messages
21
Reaction score
0

Homework Statement


An ideal gas at 300 K and 1.5x10^5 Pa is contained in a piston cylinder device. The piston is 0.1m above the base of the cylinder, has a diameter of 0.15 m and is allowed to move freely. Heat is transferred to the cylinder causing the piston to move a distance of 0.05 m. The gas specific heat at constant volume is 718 J/kg·K and the gas constant is 287 J/kg·K.
(a) What is the work done by the gas and the gas pressure at the end of the process? (b) Calculate the final gas temperature and the amount of heat transferred to the gas during the whole
process. (Ans: (a) -132.54J, 1.5x10^5 Pa, (b) 450K, 464.26 J )

Homework Equations


The Attempt at a Solution


a)
The pressure is constant = 1.5x105 Pa I have no idea why...
WD by gas = (1.5x10^5)(0.1-0.05)pi(0.075^2)=132.54J <---- It should be positive right ? The answer shows negative.
b)
T=300(0.15A)/(0.1A)=450K , A is the area.
OK the main problem starts here. After a few tries i got
n=PV/(T287) <-------------so this one i should take 287 instead of the usual 8.31? how come ? I know the qn says gas contant is 287 but why did it change ?
deltaU=(fR/2)(ndeltaT) <---------- Is this eqn right ? f is the degree of freedom since we aren't told what type of gas this is.
I took fR/2 as the specific heat at constant vol( correct?)
and obtain deltaU=331.57
For energy conservation the heat supplied must cause the increase in U by 331.57 and must also provide for the WD by gas. Q=331.57+132.54=464.11 Actual answer is 464.26.
I cannot grasp this properly. This is a constant pressure process. So why do we use constant vol specific heat?
 
Physics news on Phys.org
Work should be positive.

Pressure is constant because problem says the piston is allowed to move freely. The portion above the piston is open to the atmosphere.

One uses constant pressure specific heat when one deals with enthalpy. You are dealing with internal energy so constant volume specific heat is used.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top