O 1.66 points I Previous Answers SerCP10 10, PO58 My Notes Before beginning a lo

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O 1.66 points I Previous Answers SerCP10 10, PO58 My Notes Before beginning a long trip on a hot day, a driver inflates an automobile tire to a gauge pressure of 1.64 atm at 300 K. At the end of the trip, the gauge pressure has increased to 2.14 atm. (a) Assuming the volume has remained constant, what is the temperature of the air inside the tire? 391.463 Your response is within 10% of the correct value. This may be due to roundoff error, or you could have a mistake in your calculation. Carry out all intermediate results to at least four-digit accuracy to minimize roundoff error K (b) what percentage of the original mass of air in the tire should be released so the pressure returns to its original value? Assume the temperature remains at the value found in part (a) and the volume of the tire remains constant as air is released. 23.40 Your response differs from the correct answer by more than 10%. Double check your calculations.% Need Help? L Read Submit Answer Save Progress 1.1/1.66 points Previous Answers SerCP10 11 P005.

Explanation / Answer

Answer:-

Given Data:-

Pi =1.64 atm

Ti =300K

Pf =2.14 atm

Part-a)

We have to calcualte the temperature inside the tire.

By using Ideal Gas Law.

PV= nRT

For Pressure and temperature formula can be rewrite

P/T = nRV .

We consider here the changes in only Pressure and temperature.

So Pi/Ti = Pf/Tf

Temperature inside the tire Tf = Pf*Ti /Pi

Plug in the values. (assume atmospheric Presure P =1 atm )

Tf = (2.14+1) atm *300K / (1.64+1) atm

Tf = 356.8181 K. (Temperature inisde the tire) .

Part-b)

By Using Formula

PV =nRT.

P/n =T

Pi/Ti =ni = 1.64/300 = 0.0054

Pf/Tf=nf = 2.14/356.8181 = 0.00599

Percentage = 0.0054/0.00599 = 0.91

so Air mass released = 100% -91% = 9%.

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