The tire on a bicycle is
filled with air to a gauge pressure of 647 kPa at 20◦C. What is the gauge
pressure in the tire after a ride on a hot day when the tire air tempera- ture
is 69◦C? (Assume constant volume and a constant atmospheric pressure of 101.3
kPa.)
Answer in units of kPa.
Answer in units of kPa.
Sol:
Here the formulae used is:
Here the formulae used is:
P1V1=n1RT1
P2V2=n2RT2
P2V2=n2RT2
Where P1=po+pg
P1=101.3kPa + 647kPa= 748.3kPa
n1=n2
also given that V1=V2.
Hence, P1/T1=nR/V=P2/T2
P1/P2=T1/T2
or
P2=(P1*T2)/T1 = (748.3 *(273+69))/(273+20)
P2=748.3kPa(342K/293K)= 748.3kPa * (1.167)=873.44kPa
P1=101.3kPa + 647kPa= 748.3kPa
n1=n2
also given that V1=V2.
Hence, P1/T1=nR/V=P2/T2
P1/P2=T1/T2
or
P2=(P1*T2)/T1 = (748.3 *(273+69))/(273+20)
P2=748.3kPa(342K/293K)= 748.3kPa * (1.167)=873.44kPa
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