A tank containing 200 L of hydrogen gas at 0.0 Celsius is kept at 10 kPa. The pressure is raised to 95C, and the volume is decre
1 answer:
Answer:
The new pressure of the gas is 15.40 kPa.
Explanation:
Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases. Mathematically this law indicates that the quotient between pressure and temperature is constant:
On the other hand, Boyle's law says that the volume occupied by a certain gaseous mass at constant temperature is inversely proportional to the pressure. This law is expressed mathematically as:
P*V=k
Finally, Charles's law indicates that as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases. Mathematically, this law says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:
Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:
Studying an initial state 1 and a final state 2, it is fulfilled:
In this case:
- P1= 10 kPa
- V1= 200 L
- T1= 0 C= 273 K
- P2=?
- V2= 175 L
- T2= 95 C= 368 K
Replacing:
Solving:
P2= 15.40 kPa
<u><em>The new pressure of the gas is 15.40 kPa.</em></u>
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A) 320 count/min
B) 40 count/min
C) 80 count/min, 11400 years
Explanation:
A)
The activity of a radioactive sample is the number of decays per second in the sample.
The activity of a sample is therefore directly proportional to the number of nuclei in the sample:
where A is the activity and N the number of nuclei.
As a consequence, since the number of nuclei is proportional to the mass of the sample, the activity is also directly proportional to the mass of the sample:
where m is the mass of the sample.
In this problem:
- When the mass is , the activity is
- When the mass is , the activity is
So we can find A2 by using the rule of three:
B)
The equation describing the activity of a radioactive sample as a function of time is:
(1)
where
is the initial activity at time t = 0
t is the time
is the decay constant, which gives the probability of decay
The decay constant can be found using the equation
where is the half-life, which is the amount of time it takes for the radioactive sample to halve its activity.
In this problem, carbon-14 has half-life of
So its decay constant is
We also know that the tree died
t = 17,100 years ago
and that the initial activity was
(value calculated in part A, corresponding to a mass of 20 g)
So, substituting into eq(1), we find the new activity:
C)
We know that a sample of living wood has an activity of
per 1 g of mass.
Here we have 5 g of mass, therefore the activity of the sample when it was living was:
Moreover, here we have a sample of 5 g, with current activity of : it means that its activity per gram of mass is
We know that the activity halves after every half-life: Here the activity has became 1/4 of the original value, this means that 2 half-lives have passed, because:
- After 1 half-life, the activity drops from 16 count/min to 8 count/min
- After 2 half-lives, the activity dropd to 4 count/min
So the age of the wood is equal to 2 half-lives, which is: