Question

On a cold winter morning when the temperature is –13°C, the air pressure in an automobile tire is 1.5 atm. If the volume does not change, what is the pressure after the tire has warmed to 15°C? i need help idk the anwser. and there is like two more questions that i also need help if you could help thanks

Answer 1

the choices are
A.3.0 atm
B.–1.5 atm
C.19.5 atm
D.1.7 atm

Answer 2

Pv=NrT are you familar with this equation?

Answer 3

nope i havent done chemistry in a while this is a credit retreval class

Answer 4

i'm guessing you have to use one of the gas laws.

Answer 5

yes i think

Answer 6

it doesnt say

Answer 7

Ok, since the volume is remaining constant, use the gay-lussac's law.

which is:

\[\frac{ P1 }{ T1 } = \frac{ P2 }{ T2 }\]

where P1 is initial pressure, T1 is initial temperature, P2 is final pressure, and T2 is final temperature.

Answer 8

i dont understand how to use that formula can explain further please

Answer 9

We have the following list of known variables:

T1(initial temperature) is -13C or 260.15 kelvins
T2(final temperature) is 15C or 288.15 kelvins
P1(initial pressure) is 1.5 atm
P2(final pressure) is the variable we want to find.

Using the gay-lussac's law, we isolate P2 from the equation.

\[P2 = T2\frac{ P1 }{ T1 }\]

Plug in the known variables to solve for the final pressure of the tire.

(note: you need to use the units kelvin for the temperature when you are solving for P2).

Answer 10

Yes, coolsday has got the write idea. For kelvin you add 273 to degrees btw

Answer 11

so do i take 1.66 and add it to 273?

Answer 12

no no no. 1.66 atm is is your final pressure.

add 273 to your temperatures to convert them to kelvins

For example,

-13C + 273 = 260 Kelvins

Answer 13

oh okay so for the other temperature it will be 288 Kelvin

Answer 14

since its 15+273

Answer 15

yes

Answer 16

and then what do you do witht the two temperatures?:/

Answer 17

do you add them to something?

Answer 18

To use the equation to find P2, you have to convert the temperatures in to kelvins in order for the final answer to be correct.

Answer 19

You use the two temperature, after converting them into kelvins, to find the final pressure (P2)

Answer 20

but so you add both temperatures together or to 1.66?

Answer 21

srry chem is really hard for me im trying to understand

Answer 22

Dont overthink the question to much. You know pressure 1 and temperture one. You also know temp 2. So P1/T1= What? It = 1.5atm/ 260K.
This gives us 5.8e-3= P2/ T2. And you also know T2 = 288 Kelvin.
So multiply out, 288 K x 5.8e-3= P2
P2= 1.7 atm

Answer 23

The equation used was:

\[P2 = T2\frac{ P1 }{ T1 }\]

As you can see, to solve for P2(final pressure), you need T2, P1, and T1

However, in order for you to get the right answer, T2 and T1 has to be converted into kelvins.

Once you convert T1 and T2 into kelvins, plug in T1, T2, and P1 into that equation to solve for P2.

Answer 24

thanks

Answer 25

no problem.

Answer 26

oh hey can you help me with this next question its the only other question i have

Answer 27

Ok, i will try

Answer 28

this is the question The mass of 322 mL of hydrogen gas at 4.0 atm and –73°C is 1.21 grams. What is the density of hydrogen gas at STP?

Answer 29

A.6.88 × 10–4 g/mL

Answer 30

B.1.45 × 103 g/mL

Answer 31

C.1.28 × 10–3 g/mL

Answer 32

D.4.87 X 101

Answer 33

STP has a pressure of 1atm and a temperature of 0 Celsius.

The variables given are:

m=1.21 g
v=322 ml
p=4.0 atm
t=200 k
n=?

use the ideal gas law which is:

\[pv=nrt\]

Okay, since you want to find the density of the hydrogen gas, you will have to manipulate the ideal gas law.

We know that density is mass / volume or \[d = \frac{ m }{ v }\]

We also know that mass is molar mass x moles or \[m = Mn\]

So, \[d = \frac{ Mn }{ v }\]

Now take the ideal gas law equation again and isolate it for n

\[n = \frac{ pv }{ rt }\]

Take \[d = \frac{ Mn }{ v }\]
and substitute n with \[n = \frac{ pv }{ rt }\]


Now we get


d = (Mpv/rt)/v

d = Mpv/rtv pr d = Mp/rt

Use the final equation \[d = Mp/rt\] to find the density

M is the molar mass of hydrogen gas
p is the pressure of the hydrogen gas
r is the gas constant (0.082057 L atm mol^-1 K^-1
t is the temperature of the hydrogen gas in kelvins

Answer 34

the solution you will get will be expressed in g/L

So you will have to convert it to g/ml

Answer 35

Sorry for taking so long, but it has been a while since i worked with gas laws :)