Finding R and a from pV measurements at different temperatures

[Noor123]

Homework Statement

Suppose that you measured the product pV of 1 mol of a dilute gas and found
that pV = 22.98 L atm at 0°C and 31.18 L atm at 100°C. Assume that the perfect gas law
is valid, with T = t(°C) + a, and that the value of R is not known.

Determine R and a from the measurements provided. Express your result for R in J K

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Homework Equations

PV = nRT
T = t(°C) + a

The Attempt at a Solution

Ok I answered the first part:
R = slope = (31.18 - 22.98 / 100 - 0 )
= 0.082 L atm C

what is a ??how do I find it??

 

[Borek]

What temperature scale should be used in the ideal gas equation?

 

[Noor123]

What do you mean?

 

[Borek]

What temperature scales do you know?

 

[Noor123]

ohh well we usually do it in Kelvin, but the question asks for Celsius, so I guess we don't have to convert anything, no?

 

[Borek]

How do you convert between C and K? What is "a"?

Note, your R value looks correct, but you are just lucky - "a" canceled out.

 

[Noor123]

C= value of Temp in C + 273.15 = value in K

I have no idea what "a" is.

I don't understand the meaning behind the given equation: T = t(°C) + a

What's t(°C) ? and how are we supposed to use it to determine "a"?

I'm so lost. :(

 

[Borek]

It is so simpe that you will feel ashamed once you will get it Compare two equations that you have wrote:

temp in K = temp in C + 273.15

T = t(°C) + a

Note, that T means absolute temperature (Kelvins), while t(°C) means temperature in Celsius degrees. Do you see that it is the same equation?

 

[Noor123]

haha ok, so a is 273.15

so you're basically saying,

PV = n R (t(°C) + a)

1) 22.98 = nRa
2) 31.18 = nR(100 + a)

If we do 2-1

31.18 - 22.98 = 100Rn + nRa - nRa
8.20 = 100R (1 mol)
R = 0.0820 L atm C

and if we plug back into one of the equations,
we get 22.98 L atm= (1mol) (0.0820L atm C)a

a = 280.24 C ?
is this our right unit?