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The balloon with larger capacity (1800 ml) will burst first.
(Here, it is assumed, balloon wall materials and gas inside spherical balloon (air) are same for both cases. And, balloon wall is thin (r/T>10), so thin walled spherical pressure vessel formula can be applied.)
Equations for determining tangential tensile stress (hoop) developed at thin balloon wall:
σ=Pr/2t.........(1)
(radial stress at inner wall ‘–P’ will be very less compared to hoop stress, so neglected.)
Gas law Equations in terms of density and specific R_specific (R_specific= R/M):
P=ρX R_specificXT.........(2)
(density, ρ and specific gas constant, R_specific is intrinsic properties of a gas, so it will be same for the same gases present in both balloons.)
From equation (1) & (2) and by partial differentiation,
dσ = (ρ X R_specific / 2t) X r X dT......(3)
so, rate of balloon wall’s stress increase is proportional to r (radius of spherical balloon).
So, balloon with larger capacity (greater volume means larger radius) will experience greater stress increase with increasing temperature. And ultimate stress at larger blloon will be crossed earlier than smaller balloon. So, larger balloon will burst first.
Here,
σ= Stress at balloon wall
P =Pressure inside balloon
r=Radius of spherical balloon
t= Thickness of balloon wall
ρ=Density of gas inside balloon
T= Temperature of gas inside balloon
R=Ideal or universal gas constant
M=Molar mass of gas inside balloon
R_specific=R/M=Specific gas constant of gas inside balloon
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