The tyre pressure of a car was found to be 150cmHg in the morning at a temperature of 27°C. Hard driving in the afternoon raised the temperature of the tyres to 57°C. The tyre pressure had by the afternoon was?
increased by 15.0cm Hg
decreased by 15.0cm Hg
increased by 22.50cm
decreased by 22.50cm Hg
remained constant
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Discussions (9)

p1= 150cmHg
T1=27°c+273=300k
p2=?
T2= 57°c+273=300k
using pressure law
P1/T1=P2/T2
150/300=P2/330
therefore
P2=150×330÷300
P2=165cmhg.
(remember P2 increase oo)
checking the options
P2-P1
165-150=15
15(increased) .
therefore A is correct

I think my school should review the answer because the answer is not provided in the options

Addition to my first error report.
150mmHG is measured overpressure over the ambiënt pressure of 76cmHg.
So right calc : 357/300 x ( 150+76) -76= 192.9 cm Hg, so 42.9 cmHg rising.
Assuming vollume of tyre dont rise by the higher pressure, wich is about right at an already pressurised tyre.

Temp was raised by 57 degr C. So became 27+57 = 84 degr C.
In the answer 57 degr C is used.
So 273+27= 300 degr K.goes to 357 degr K. Gives 357/300 x 150cmHg = 178cm Hg.
So 28cm rising of pressure.
So no answer is right.


