Question

The pressure in a bicycle tire is __55.0__psi at __30.0__˚C in Phoenix. You take the bicycle up to Flagstaff, where the temperature is _5.0___˚C. What is the pressure, in kPa, in the tire?(Volume and Amount of moles is held constant)

Answer 1

The pressure in the tire at 5.0 °C is 347.91 kPa

Step-by-step explanation:

Step 1: Data given

The pressure in a bicycle tire is 55.0 psi

Temperature = 30.0 °C = 303 K

Temperature decreases to 5.0 °C = 278 K

Volume and Amount of moles are held constant

Step 2: Calculate the pressure at the new temperature

P1/T1 = P2 / T2

⇒with P1 = the initial pressure of the bicycle tire is 55.0 psi

⇒with T1 = the initial temeprature = 303 K

⇒with P2 = the pressure at the new temperature

⇒with T2 = the decreased temperature = 278 K

55.0 psi / 303 K = P2 / 278 K

P2 = (55.0 psi / 303 K) * 278 K

P2 = 50.46 psi

Step 3: Convert pressure from psi to kPa

50.46 psi = 50.46 * 6.895 = 347.91 kPa

The pressure in the tire at 5.0 °C is 347.91 kPa

Answer 2

`p_2=347.9kPa`

Step-by-step explanation:

Hello,

In this case, we use the Gay-Lussac's law which allows us to understand a gas' pressure-temperature behavior as a directly proportional relationship:

`(p_1)/(T_1)= (p_2)/(T_2)`

Whereas it is convenient to use the pressure in kPa and the temperature in kelvins in order to compute the required resulting pressure, therefore:

`p_1=55.0psi*(6.89476kPa)/(1psi) =379.2kPa\\T_1=30.0+273.15=303.15K\\T_2=5.0+273.15=278.15K`

Thus, we obtain:

`p_2= (p_1T_2)/(T_1)=(379.2kPa*278.15K)/(303.15K)\\ \\p_2=347.9kPa`

Best regards.