Combined Gas Law Homework: Find Final Temp (K)

[Gyro]

Homework Statement

250L of air initially at a pressure of 250Pa and a temperature of 50 degrees F is heated until the gas is at a pressure of 500 Pa and occupies 500L What is the final temperature of the air* in Kelvin. *Assume air behaves like an ideal gas.

Homework Equations

$$\frac {P_{1}V_{1}}{T_{1}}=\frac {P_{2}V_{2}}{T_{2}}$$

The Attempt at a Solution

First I converted to Kelvin.
$$T_{1} = \frac {50 - 32}{1.8} + 273 = 283 K$$
Then I used:
$$T_{2} = \frac {T_{1}P_{2}V_{2}}{P_{1}V_{1}}$$
$$T_{2} = \frac {(283K)(500Pa)(500L)}{(250Pa)(250L)}=1132K$$
But a temperature of 1132K seems high to me, is this wrong?
Thanks in advance.

 

[thepatient]

You must use atmospheres in the calculations, not Pascals. But you are dividing by pascals though... Seems right. It heats up until the volume is twice as much and you get twice the pressure as well. That is a lot of heat.

 

[Gyro]

You must use atmospheres in the calculations, not Pascals. But you are dividing by pascals though... Seems right.

huh? In your first sentence you say I did it wrong by using Pa and not atmospheres. Then in the next sentence you say it's ok though because I divided by Pa and that it seems right.
I'm confused. Is the final temp correct?
Thanks for your reply.

 

[thepatient]

Lol yes. It seems fine. Normally when calculating using the ideal gas laws you would use atm only, but since in this case you are multiplying by units in pascal and dividing by pascals, you get the same answer. I suggest always using the standard units for the gas laws though. Not everything would work out the same way. If you use temperature to stay at celsius or Fahrenheit for example, you won't get the same value.

1 atm = 101325 Pa

So if you have like 1 atm as an initial pressure and 2 atm as a final pressure, the ratio will be 2atm/1atm, which will be the same ratio as 202650Pa/101325Pa.

1 degree celsius = (1 + 273.15) kelvin.

If you have as initial temperature to be 1 degree celsius and final to be 2 degrees celsius, the ratio of 2*C/1*c isn't the same as 275.15*K/274.15*k

 

[Gyro]

... I suggest always using the standard units for the gas laws though. Not everything would work out the same way. If you use temperature to stay at celsius or Fahrenheit for example, you won't get the same value.

I see. So using Kelvin is necessary then, because if I use Celsius or Fahrenheit, I would get an incorrect answer. I think that's what you mean by using standard units. Celsius and Fahrenheit are not for Gas Law questions. Is that what you mean?
I'm not that good at chemistry and thermo. Thanks for your help.

 

[Borek]

Normally when calculating using the ideal gas laws you would use atm only, but since in this case you are multiplying by units in pascal and dividing by pascals, you get the same answer.

You are wrong about using atm only, then you are right saying that in the end one gets the same answer no matter what pressure units are used, as long as they are used consistently.

In fact you can use any volume and pressure units you like (including Torr and PSI), just select appropriate ideal gas constant value (in appropriate units). See table in wikipedia (http://en.wikipedia.org/wiki/Ideal_gas_constant).

If anything, the most hardcore correct pressure units is Pa, as it is an SI unit.

As for the temperature - any absolute scale will work OK (absolute meaning it has zero at absolute zero). We usually use Kelvin, but Rankine would work OK as well (after modifying R value).

It doesn't mean you have to use these fancy approaches, just be aware when and why they are correct.

 

[Gyro]

... If anything, the most hardcore correct pressure units is Pa, as it is an SI unit.

As for the temperature - any absolute scale will work OK (absolute meaning it has zero at absolute zero). We usually use Kelvin, but Rankine would work OK as well (after modifying R value)...

Thank you for clearing that up, Borek.

 

[donutz610]

Convert Pascals to Atmospheres using 1atm equals 101325pa and that should decrease your answer by quite a bit

 

[Borek]

Convert Pascals to Atmospheres using 1atm equals 101325pa and that should decrease your answer by quite a bit

No, as already explained it won't change the answer.