Ideal Gas Exercise: Homework Statement & Solution

AI Thread Summary
A cylinder with a frictionless piston contains gas initially at 300K and 6.0 * 10-3 m3, which is heated to 400K. The initial energy was calculated as 900J, and the second energy as 1200J, leading to a work calculation of 300J. However, the correct work done by the gas is 200J, as the change in internal energy does not account for heat transfer. The correct approach involves using the equation w = -PextΔV, where Pext is the constant external pressure, simplifying the calculation. Understanding this integral approach clarifies the relationship between work, heat transfer, and internal energy changes.
Perrin
Messages
14
Reaction score
0

Homework Statement



A cylinder with a frictionless piston is placed horizontally in an atmosphere pressure 1 * 105 N/m2. A gas in the cylinder is initially at a temperature of 300K with a volume of 6.0 * 10-3 m3. Then the gas is heated slowly to 400K. How much work is done by the gas in the process?

Homework Equations



Ideal Gas equations:

p*V/T = constant

p*V = NKT = nRT

Ek (average) = (3/2) * kT

Ek (total) = (3/2) * NkT = (3/2)pV

The Attempt at a Solution



At first I calculated the initial energy as:
E1 = (3/2)pV = (3/2)*(1 * 105) * (6.0 * 10-3) = 900J

Then, assuming the pressure is constant, I said:

pV1/T1 = pV2/T2

V2 = V1*T2 / T1

V2 = 6.0 * 10-3 * 400 / 300 = 8*10-3

Thus, the second energy:

E2 = (3/2)*105*8*10-3 = 1200J

And the work:
W = E2 - E1 = 1200 - 900 = 300J.

Now, in the answer to the question, it says the work is not 300J, but 200J.
Can someone enlighten me about my mistake?

Thanks.
 
Physics news on Phys.org
You calculated the change in internal energy of the gas (ΔE). ΔE = w + q, so you are ignoring q, the amount of heat transferred during the process. You should use this equation for work instead:

w = -\int_{v_1}^{v_2}{P_{ext}dV}
 
So, I could also calculate:

w = ΔE - q ?

And could you please explain to me how to use the equation you wrote? What's Pext? How do you integrate it? Sorry, I'm not very good with integrals...

Thanks for the quick reply!
 
Pext is the external pressure. If the external pressure stays constant during your process, P does not vary with V and you can pull it out of the integral to get the simpler relation w = -PextΔV. If P varies as the volume changes, you have to do the integration by finding an expression for P in terms of V.
 
Thanks, I finally understand :D
 
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