MCQ: Ideal Gas Flow Out of Balloon with Temp Increase

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SUMMARY

The discussion centers on calculating the fraction of gas flow out of a balloon when the temperature increases from 1000°C to 1020°C, while maintaining constant volume. The initial calculation suggests a flow fraction of 2/373 based on the ideal gas law (PV=nRT), but the correct answer is 2/375, which accounts for the increase in mass as temperature rises. The distinction lies in recognizing that while volume remains constant, the number of moles of gas changes with temperature, leading to different interpretations of the problem.

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Homework Statement


A balloon with a constant volume is in 1000C. The fraction of gas flow out of balloon when temperature increased by 20C is nearly equal to
(consider pressure and the gas in balloon is ideal)

a. 2/373
b. 2/375
c. 2/100
d. 373/375
e. 100/102

Homework Equations


PV=nRT
P-pressure
V-volume
n-number of mols
R-universal gas constant
T-temperature

The Attempt at a Solution


In these conditions P,n & R is constant
∴ V α T
V1/V2=T1/T2
V1/V2=375/373

the amount flow out = 375/373-1 = 2/373

but the answer given as as 2/375
the book i get this considered the mass is increasing when temperature increases

in book that was like

PV=(m/M)RT
P,V,M and constant (m-mass of gas, M-molar mass)

m α 1/T
m1T1=m2T2
m1/m2=373/375

the mass flow out = 1-373/375=2/375

which is right
 
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kalupahana said:
A balloon with a constant volume is in 1000C. The fraction of gas flow out of balloon when temperature increased by 20C is nearly equal to
(consider pressure and the gas in balloon is ideal)


The Attempt at a Solution


In these conditions P,n & R is constant
∴ V α T
V1/V2=T1/T2
V1/V2=375/373...

the amount flow out = 375/373-1 = 2/373

but the answer given as as 2/375

The volume of the balloon is constant, but n, the amount of gas in it is different. The pressure is the same at both temperature. You have the equations

PV=n1 R T1
and
PV=n2 R T2.

ehild
 

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