How Does Heating Affect the Internal Energy of Gas in a Piston-Cylinder System?

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Heating the gas in a piston-cylinder system increases its internal energy, which can be calculated using the first law of thermodynamics, where the change in internal energy (delta U) equals heat added (Q) minus work done by the gas (W). The heat added is determined using the heat capacity at constant pressure, resulting in Q = 50000 J for a temperature increase from 300K to 400K. Work done by the gas is calculated using the pressure and change in volume, but the correct pressure must be determined from the system's conditions. The user initially calculated W as 456500 J, leading to an incorrect delta U value. Further clarification on the pressure and volume calculations is needed to resolve the discrepancies in the internal energy change.
huskydc
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A cylinder (cross section is 0.2m2) with a free moving piston is filled with gas. The piston is attached to a heavy weight W = 10000N. Outside the cylinder, the air is at 300K and 1atm. Initially the gas is at 300K, then it is heated to 400K. The heat capacity of the gas under the constant pressure is 500J/K.
If the length of the gas in the cylinder l increases by 20cm during the heating, find the change in the internal energy of the gas in Joule J.



i figured...heat is being transferred to the gas, and work is done by the gas, so both Q and W should be postive...

delta U = Q - W

i'm not really sure how to start this...but i tried the following:

C = Q/ delta T

500 = Q / (400-300)

Q = 50000 J

W = P delta V (i'm not sure if the P is of the system or the surrounding)

but with the problem, I've been given a clue that says this:

P system = (P surrounding - W force)/cross area...

and I came up with 456500 J...

Q- W

50000 - 456500 = delta U...apparently the answer didn't work, any clues?
 

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:frown: :frown: can i even use the equation: delta U = 1.5 n R delta T
but that would include finding out n...which i don't have enough info to calculate...any one help??
i'm lost... =(
 
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Thread 'Variable mass system : water sprayed into a moving container'

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}...
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