Calculate the ratio of new volume to old volume

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To calculate the ratio of new volume to old volume of a car tire under constant pressure, the general gas equation is applied. Given the initial temperature of 23°C (296 K) and final temperature of 34°C (307 K), the relationship derived from Charles' Law indicates that the volume ratio is equivalent to the temperature ratio. The calculation results in a volume ratio of 1.04 when comparing the final and initial temperatures. This confirms the understanding that under constant pressure, the volume of a gas is directly proportional to its temperature. The conclusion affirms the correct application of gas laws in solving the problem.
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Homework Statement


A car tyre has initial temperature 23°C and inital pressure 2*105 Pa
After heating in the sun, temperature is now 34°C and final pressure is 2.1*105Pa

Question asks, Calculate the ratio of new volume to old volume (V1/V2) if the pressure is held constant during the temperature rise stated above.

Homework Equations


General Gas Equation

(P1V1)/T1= (P2V2)/T2

The Attempt at a Solution


Using the general gas Equation I worked out the final pressure of 2.1*105Pa which was not orignially given.

Now for the ratio my idea is as follows

If the pressure is held constant, then V1/T1= V2/T2 which is a statement of charles Law.

I'm not sure how to find the ratio but perhaps if the ratio of new to old volume would be the same as the ratio between new and old temperature?

SO I would convert the celciusto kelvin by adding 273 and obtain 296 Kelvin initial temperature, and 307 K final temperature

Then I would do this 307/296 and that would give me 1.04 to two significant figures as the ratio.

Is this correct? I would like some assistance.
 
Last edited:
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Richie Smash said:
perhaps if the ratio of new to old volume would be the same as the ratio between new and old temperature?
That is what Charles' Law says. Just rearrange the equation to that form.
Richie Smash said:
Is this correct?
Yes.
 
Oh ok thanks alot! I shall re arrange the equation for further clarification
 
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