A spherical balloon filled with heliumm is initially at 300 K and 100 kPa with a

Question

A spherical balloon filled with heliumm is initially at 300 K and 100 kPa with a diameter of 5 cm. The balloon is inflated until it contains 4 times its initial mass. During the inflating, heat is being transferred to the balloon such that the final temperature of the Helium is 450 K. Due to the elasticity of the balloon, the pressure remains proportional to the diameter squared at all times. Determine the work (kJ) and heat transfer (kJ) associated with this process.

Note: cp/R (Helium) = 2.5 cv=cp-R

T1= 300K T2= 450K

P1= 100 kPa P2= 204.77 kPa

V1= V2= 1.973-4m^3

m1= m2= 4.20e-5 kg

D= 0.05m

Explanation / Answer

P1 = 100000 = k*0.05^2

k = 4*10^7

P2 = 4*10^7*(4^(1/3)*0.05)^2 = 251.98kPa

P= k*d^2

P=k*(6V/3.142)^(1/3)=4*10^7*1.24*v^(1/3)= 4.96*10^7*v^(1/3)

w = int. PdV = 3.72*10^7*(V2^(4/3)- V1^(4/3))

V2 = 3.142*(4^(1/3)*0.05)^3/6 = 0.0002618333

V1 = 3.142*(0.05)^3/6 = 0.00006545833

W = 3.72*10^7*(0.00001675071-0.00000263807) = 525J work done

Q = U + w = nCpdT + w

n = 10^5*0.00006545833/(8.314*300) = 0.00262

Q = 0.00262*2.5*8.314*150 + 525 = 533.168 J heat

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