0:00
Hi there.
00:01
So for this problem, we're told that a small amount of water and its saturated paper are enclosed in a vessel at the temperature that is given.
00:10
This is the, well, we are given the temperature and that is equal to 100 celsius degrees.
00:18
We know that to pass this into kelvin, we just need to add to this value to 173.
00:24
So the value that we obtained from this is 373 kelvin.
00:28
And so the question is how much will the mass of the saturated paper increase if the temperature of the gas of the system up by a change in temperature that is equal to 1 .5 kelvin.
00:49
And we are also told that to assume that the paper is an ideal gas and the specific volume of water is negligible as compared to that of paper.
00:59
Now with that said, we start with the expression that the derivative of the pressure with respect to the temperature is approximately the molar mass times the heat, and this divided by the gas constant times the temperature to the square times the pressure.
01:22
Now, in this case, if we solve, we can solve for the derivative of the, um, differential in the pressure divided by the pressure is equal to just simply the molar mass times this, this divided by the gas constant, times the temperature, and this times the differential in the temperature divided by the temperature because we know that this part in here will just yell the previous expression where the temperature is to the square.
02:00
Now, with that said, we are going to use now the ideal gas law.
02:11
That states that the product between the pressure and the volume is equal to the mass divided by the molar mass, and these times the gas comes down r times that the imperative.
02:24
Now, what i'm going to do is to apply the logarithm, the neferian logarithm, to both sides of this expression.
02:34
And using properties of the neferian logarithm, we are going to obtain the follow.
02:40
The apparent logarithm of the pressure times the neparian logarithm of the volume and then this is equal to the neparion logarithm of the mass and this plus the neparion logarithm of the temperature plus the neparion logarithm i'm going to consider this term as one because that is going to be a constant and now i'm going to to derive both sides of this expression.
03:13
So we know that this is a constant because we are told the volume is a constant and also this is a constant.
03:19
Then we will have that this is the derivative of the pressure with respect divided by the pressure or the differential and pressure because remember that the derivative of the nepearian logarithm is the inverse.
03:34
And then we will have this is just the same for mass and the same for the temperature...