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[SOLVED] Thermodynamics Question and My Answer
During a cloudy day of the summer season, a driver inflates his car tyres (max allowed pressure is 44PSI(Gauge)) to 38PSI(Gauge) at a petrol station. the temperature of the air inside one of the tires is [tex]77^oF[/tex] and volume of the tire is [tex]1ft^3[/tex]. Later, the sun comes out and the air inside the tire experience a temperature rise of [tex]36^oF[/tex]. Assuming the volume to be constant determine whether the tire will burst or not.
P(Absolute) = P(Ambient) + P (Gauge)
Conversion from F to R
Conversion from PSI to [tex]\frac{lbf}{ft^2}[/tex]
First did the relevant conversions.
Then used :-
[tex]P_{1}v = RT_{1}[/tex]
[tex]P_{2}v = RT_{2}[/tex]
Where [tex]R[/tex] = Individual Gas Constant
Where [tex]v[/tex] = Specific Volume
Took the ratios, because the volume remains constant and so does the value of R :-
[tex]\frac{P_{1}}{P_{2}} = \frac{T_{1}}{T_{2}}[/tex]
Made relevant calculations and then did reverse conversions. Got the answer to be 41.54 PSI (Gauge). Which indicates that tire doesn't bursts. Is this the correct answer ?
Homework Statement
During a cloudy day of the summer season, a driver inflates his car tyres (max allowed pressure is 44PSI(Gauge)) to 38PSI(Gauge) at a petrol station. the temperature of the air inside one of the tires is [tex]77^oF[/tex] and volume of the tire is [tex]1ft^3[/tex]. Later, the sun comes out and the air inside the tire experience a temperature rise of [tex]36^oF[/tex]. Assuming the volume to be constant determine whether the tire will burst or not.
Homework Equations
P(Absolute) = P(Ambient) + P (Gauge)
Conversion from F to R
Conversion from PSI to [tex]\frac{lbf}{ft^2}[/tex]
The Attempt at a Solution
First did the relevant conversions.
Then used :-
[tex]P_{1}v = RT_{1}[/tex]
[tex]P_{2}v = RT_{2}[/tex]
Where [tex]R[/tex] = Individual Gas Constant
Where [tex]v[/tex] = Specific Volume
Took the ratios, because the volume remains constant and so does the value of R :-
[tex]\frac{P_{1}}{P_{2}} = \frac{T_{1}}{T_{2}}[/tex]
Made relevant calculations and then did reverse conversions. Got the answer to be 41.54 PSI (Gauge). Which indicates that tire doesn't bursts. Is this the correct answer ?
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