can someone pleeease help me with this physics problem?

Question

at the start of her trip on a pleasant afternoon at 22C, a driver's car tires have a pressure of 2.9x10^5 Pa. The pressure rises by 33% during her trip. Ignoring the expansion of the tires, find the temperature of the tires at the end of her trip

** can you please show it step by step?

Answer 1

Here, air is assumed as ideal, and the container (the tyre) is considered rigid [the volume remains constants].
For this case, we can therefore use the ideal gas law equations.
PV = nRT, where P is pressure in Pa, V is volume in Litres, n is the number of moles or whatnot, R is the such and such 8.31 constant [n and R aren't going to matter, since they're going to be the same, since you're dealing with the same stuff. You're only concerned with P and T in the above equation], and T is temperature in Kelvin scale.

so, you're considering everything at state 1, and then everything at state 2.

T_1 = 22C ..which is 273 + 22 in Kelvin. ..so, T1 = 295K
P_1 is 2.9x10^5 Pa. P_2 is 33% more, so multiply it:
P_2 = 1.33x(2.9x10^5) Pa.

Since the pressure is higher, you know the temperature will have to be higher (it's the only thing that can cause the increase in Kinetic Energy that caused the pressure to increase)

the ideal gas equations work out such that

P_2/P_1 = T_2/T_1

So.. ..that would mean (and I haven't checked this math 'cause I'm lazy and it's 6am and I've not been to bed yet) that T_2 = 1.33xT_1
so T_2 = 392.35K ..so to get celsius from kelvin, subtract 273.
T_2 = 119.35C

Answer 2

let p1 and t1 be the initial pressure and temperature of the air in the tyre, and p2 and t2 those of yhe escaping air after burst. For adiabatic(sudden) expansion of air, we have
gamma(y)= 1.4 for air
and,
(t1^y)/p1(y-1) = t2 ^ y/( p2 ^ (y-1))

t1= 22 +273 = 295 K

t2 = ?

p1 = 2.9 * 10^5 Pa

p2 = (.33*p1) + p1 (there is a rise in pressure)
= 385700 Pa

substituting the values in adiabatic equation:

295 ^ 1.4 / (2.9 *10^5 )^ .4 = t2 ^ 1.4 / ( 385700 )^.4

solving the equation we get:

t2^ 1.4 = 3215 . 818
t2 = 320 .04 K
or, 320-273 = 47 C

Answer 3

Because PV = nRT for an ideal gas, T2/T1 = P2/P1 = 1.33.

Therefore, T2 = 1.33xT1. Temp must be in deg Kelvin, so T1 = 273+22 = 295K. T2 = 392K = (392-273)deg C = 119C

Answer 4

we have : PV = nRT and in this question V & n & R do not change so we can say : P1/T1 = P2/T2 (here 1 & 2 are index)
then :
1 atm = 101325 Pa ---> 290000 Pa = 2.862 atm = P1
P2 = 2.862 + 2.862*(33/100) = 3.806 atm
T1 = 22 + 273 = 295K
2.862/295 = 3.806/T2
T2 = (3.806*295)/2.862 = 392.3K = 119.3C (392.3 - 273 = 119.3)

Answer 5

assume the falling body reaches the earth at (t+2) then 2 seconds earlier accomplishing the earth, time is t: X1=0.5gt^2 X2=0.5g (t+2) ^2 X2-X1=40m (0.5gt^2+2gt+2g)-0.5gt^2=40 20t+20=40 t=a million after substitution: X2=5(3) ^2=45m top of the tower is 45m.