Pressure -- open a window on a box

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SUMMARY

The discussion revolves around calculating the time required for the pressure inside a box to equal atmospheric pressure after opening a window. Given parameters include an initial volume (V1) of 2 m³, an initial pressure (p1) of 0.5 KPa, and an atmospheric pressure (P2) of 101 KPa. The formula derived for time is time = (V1 * p1) / (P2 * Q), where Q represents the volume flow rate of air entering the box. The main challenge highlighted is determining the value of Q.

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  • Familiarity with pressure-volume relationships in gases
  • Knowledge of fluid dynamics, specifically flow rates
  • Ability to manipulate and solve algebraic equations
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  • Research the concept of flow rate (Q) in fluid dynamics
  • Study the ideal gas law and its applications
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Students studying physics, particularly those focusing on thermodynamics and fluid dynamics, as well as educators seeking to explain pressure dynamics in real-world applications.

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Homework Statement



if i have a box of:
volume V1= 2 m3
pressure p1= 0.5 KPa
temperature T1= 80° C
with a window of A=0.2 m2

if i open the window what is the time for pressure box= pressure atmosphere?
temperature atmosphere T2= 25°C
pressure atmosphere P2=101kPa

Homework Equations



Q=V3/time [m3/sec]
V3 volume air goes inside the box

V1p1=V3p3=V3p2
V3=(V1p1)/p2
them
time=V3/Q=(V1p1)/(p2Q)

but Q? i don't understand
 
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Hi opsops. Welcome to Physics Forums. :wink:

That is an interesting question you have. You made it up, did you? :smile:
 

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