Work- constant pressure, changing volume

AI Thread Summary
The work done by a gas during expansion from 1.0 L to 3.3 L at a constant pressure of 2.8 atm was initially calculated incorrectly. The pressure was converted to Pascals, resulting in 283710 Pa, and the work formula W = p(deltaV) was applied. However, the correct approach requires converting the volume to cubic meters and the pressure to kilopascals by multiplying by 101.3. After making these conversions, the calculation can be redone for an accurate result. Proper unit conversions are essential for obtaining the correct answer in physics problems.
physics1234
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Find the work done by a gas when it is expanded from a volume of 1.0 L to 3.3 L at a constant pressure of 2.8 atm.

I converted the atm to Pascals by multiplying 2.8 x 101325 to get 283710.

Then I used the formula W=p(deltaV)

so 283710 x (3.3-1.0) = 652533

I don't understand why this is the wrong answer.
 
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Nevermind, I got it :-)

Thanks anyway!
 
hi brother
so as to your answer be at joule
you must convert the volume to cubic meter m3 , and to convert the pressure to kpa not pascal , by multiplying by 101.3 , try the new answer...

sorry my english is not so good
 
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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