The volume of the Cl₂ gas is 4.15 L.
Explanation:
This problem can be easily solved using the ideal gas law.
As in ideal gas law, it is stated that the product of pressure and volume is equal to the product of number of moles with temperature with gas constant.
PV=nRT
So, in the present case, pressure P is given as 177 kPa, temperature T = 348 K and the mass of chlorine gas is given as 18 g.
So first step, we have to determine the value of 'n' using the mass of Chlorine gas given. The mass should be converted to moles by dividing the mass with molar mass of the gas.
![n = \frac{Mass}{Molar mass}](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7BMass%7D%7BMolar%20mass%7D)
As molar mass of Cl is 35.453 g/mol, then ![Molar mass of Cl_{2} = 2 * 35.453 = 70.906 g/mol](https://tex.z-dn.net/?f=Molar%20mass%20of%20Cl_%7B2%7D%20%3D%202%20%2A%2035.453%20%3D%2070.906%20g%2Fmol)
So, ![n = \frac{18 g}{70.906 g/mol} =0.254 moles](https://tex.z-dn.net/?f=n%20%3D%20%5Cfrac%7B18%20g%7D%7B70.906%20g%2Fmol%7D%20%3D0.254%20moles)
Then, the volume can be determined with R = 8.3144×10³ L Pa K⁻¹ mol⁻¹ as shown below:
![V = \frac{nRT}{P} =\frac{0.254*8.3144*10^{3} *348}{177*10^{3} } =4.15 L](https://tex.z-dn.net/?f=V%20%3D%20%5Cfrac%7BnRT%7D%7BP%7D%20%3D%5Cfrac%7B0.254%2A8.3144%2A10%5E%7B3%7D%20%2A348%7D%7B177%2A10%5E%7B3%7D%20%7D%20%3D4.15%20L)
Thus, the volume of the Cl₂ gas is 4.15 L.
Answer:
False
Explanation:
A community consists of all the organisms in an ecosystem that belong to the same species. Population density occurs when two or more organisms seek the same resource at the same time. A carrying capacity is anything that restricts the number of individuals in a population.
Answer : (C) 854.46 KPa.
Solution : Given,
Initial pressure = 400 KPa
Initial temperature = 110 K
Final temperature = 235 K
According to the Gay-Lussac's law, the absolute pressure is directly proportional to the absolute temperature at constant volume of an ideal gas.
P ∝ T
Formula used :
![\frac{P_{1}}{P_{2}}=\frac{T_{1}}{T_{2}}](https://tex.z-dn.net/?f=%5Cfrac%7BP_%7B1%7D%7D%7BP_%7B2%7D%7D%3D%5Cfrac%7BT_%7B1%7D%7D%7BT_%7B2%7D%7D)
where,
= initial pressure
= final pressure
= initial temperature
= final temperature
Now put all the values in above formula, we get
![\frac{400}{P_{2}}=\frac{110}{235}](https://tex.z-dn.net/?f=%5Cfrac%7B400%7D%7BP_%7B2%7D%7D%3D%5Cfrac%7B110%7D%7B235%7D)
By rearranging the terms, we get the value of new/final pressure.
= 854.5454 KPa
854.55 KPa