How Is Work Calculated in Isothermal Compression of Air?

AI Thread Summary
The work done in isothermal compression of air can be calculated using the integral of pressure times the change in volume, with limits set at the initial and final volumes. The specific problem involves compressing 2kg of dry air to one-tenth its volume at 15°C. It is noted that the equation does not include a variable for the mass of the air parcel, and the pressure value is also unspecified. Setting up the integral with variables for unknowns may lead to a solution that is independent of those unknowns. This approach allows for a general understanding of the work involved in the process.
icnhoj
Messages
1
Reaction score
0
1. Calculate the work done in compressing isothermally 2kg of dry air to one-tenth its volume at 15C



2. Work= the integral of pressure times the change in volume, the limits of integration being the initial and final volumes, v1 and v2.



3. The equation is very general, and doesn't have a variable for mass of the air parcel. The problem also doesn't have a value for p, so I don't know if I am supposed to have a solution with p as a variable in it...
 
Physics news on Phys.org
Just set up the integral leaving variables where you don't know something. You may find the answer doesn't depend on things you don't know.
 
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