The Ideal Gas Law and Calculating Volume at Different Temperatures

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The discussion centers on applying the Ideal Gas Law to calculate the volume of an air bubble as it rises from the bottom of a lake to the surface. The initial volume is 18 cm³ at a depth of 40 m and a temperature of 3.0°C, while the surface temperature is 30°C. Users are encouraged to clarify the relationship between pressures p1, p2, and p0, and to confirm the correct setup of the equations. One participant shared their calculation method, using the formula p1v1/T1 = p2v2/T2, and expressed confusion over their results. The thread emphasizes the importance of correctly identifying initial and final temperatures and pressures in the calculations.
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An air bubble of volume 18 cm3 is at the bottom of a lake 40 m deep where the temperature is 3.0°C. The bubble rises to the surface, which is at a temperature of 30°C. Take the temperature of the bubble to be the same as that of the surrounding water. What is the volume of the bubble just as it reaches the surface?

So, my teacher told me to use p1v1/T1 = p2v2/T2 p1=pgh+p0

however, when i did the work, the answer i got was 19.7608
So, could someone set up the equation for me in case I'm doing it wrong?
 
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How is p2 related to p1 and p0? What did you use for T1 and T2? Show us how you did the calculation.
 
the only parts that i understand is the first equation which i set up like this..

18/(3+273)=x/(30+273)
 
i solved it..thank you though
 
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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