A motorist inflates the tires of her car to a pressure of 180 kPa on a day when the temperature is -8.0° C. When she arrives at
her destination, the tire pressure has increased to 245 kPa. What is the temperature of the tires if we assume that the tires do not expand?
1 answer:
Apply Gay-Lussac's law:
P/T = const.
P = pressure, T = temperature, the quotient of P/T must stay constant.
Initial P and T values:
P = 180kPa, T = -8.0°C = 265.15K
Final P and T values:
P = 245kPa, T = ?
Set the initial and final P/T values equal to each other and solve for the final T:
180/265.15 = 245/T
T = 361K
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