at STP a certain mass of gas occupies a volume of 790 cm cube find the temperature at which the gas occupies 1000 cm cube and has a pressure of 726 mmhg
Use (P1V1/T1) = (P2V2/T2)
Remember to use kelvin for T1 and T2. STP means 22,400 cc @ 273 kelvin.
Post your work if you get stuck.
At STP the density of chlorine is 3.22 kg/cu. meter. What is the weigh of this gas is contained in a flask of 100 cubic centimeters at 24 deg celcius and 100 kPa?
To solve this problem, we can use the ideal gas law equation, which is:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles of gas
R = gas constant
T = temperature in Kelvin
First, we need to calculate the initial condition, which is at STP (standard temperature and pressure). At STP, the pressure is 1 atmosphere (atm), the volume is 22.4 L (which can be converted to cm^3 by multiplying it by 1000), and the temperature is 273.15 Kelvin.
P1 = 1 atm
V1 = 22.4 L = 22,400 cm^3
T1 = 273.15 K
Now, we can calculate the initial number of moles of gas (n1) using the ideal gas law:
P1V1 = n1RT1
n1 = (P1V1) / (RT1)
Since the gas constant R = 0.0821 L·atm/mol·K, we need to convert the volume and pressure to liters and atmospheres respectively:
P1 = 1 atm
V1 = 22,400 cm^3 / 1000 = 22.4 L
T1 = 273.15 K
Now we have all the values needed to calculate n1:
n1 = (1 atm * 22.4 L) / (0.0821 L·atm/mol·K * 273.15 K)
n1 ≈ 1 mol
Now, we can use the ideal gas law equation to find the temperature (T2) at the new condition:
P2 = 726 mmHg = 726 mmHg * (1 atm / 760 mmHg) ≈ 0.955 atm
V2 = 1000 cm^3 / 1000 = 1 L
Using the same amount of moles (n1) and the new volume and pressure, we can solve for T2:
P2V2 = n1RT2
T2 = (P2V2) / (n1R)
Substituting the values:
T2 = (0.955 atm * 1 L) / (1 mol * 0.0821 L·atm/mol·K)
T2 ≈ 11.65 K
Therefore, at a volume of 1000 cm^3 and a pressure of 726 mmHg, the temperature of the gas is approximately 11.65 Kelvin.
To find the temperature at which a gas occupies a given volume and has a certain pressure, we can use the Ideal Gas Law equation: PV = nRT, where P is the pressure, V is the volume, n is the moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.
Since the question mentions STP (Standard Temperature and Pressure), we know that at STP, the temperature is 273.15 Kelvin and the pressure is 1 atmosphere (which is equal to 760 mmHg).
Let's solve the problem step by step:
1. Convert the initial volume from cm^3 to liters:
- 790 cm^3 / 1000 cm^3/L = 0.79 L
2. Convert the pressure from mmHg to atm:
- 726 mmHg / 760 mmHg/atm = 0.9553 atm
3. Set up the initial condition in the Ideal Gas Law equation:
- P1V1 = nRT1
4. Rearrange the equation to solve for n:
- n = (P1V1) / (RT1)
5. Convert the initial temperature from degrees Celsius to Kelvin:
- T1 = 0 degrees Celsius + 273.15 = 273.15 K
6. Set up the final condition in the Ideal Gas Law equation:
- P2V2 = nRT2
7. Rearrange the equation to solve for T2:
- T2 = (P2V2) / (nR)
Now, substitute the known values into the equation:
- P1 = 1 atm
- V1 = 0.79 L
- T1 = 273.15 K
- P2 = 0.9553 atm (converted from 726 mmHg)
- V2 = 1 L (converted from 1000 cm^3)
- n = (P1V1) / (RT1)
Substitute these values into the equation and solve for n:
- n = (1 atm * 0.79 L) / (0.0821 atm·L/mol·K * 273.15 K)
- n = 0.027 mol (rounded to 3 decimal places)
Now, substitute the values into the final equation:
- T2 = (0.9553 atm * 1 L) / (0.027 mol * 0.0821 atm·L/mol·K)
- T2 = 429.56 K (rounded to 2 decimal places)
Therefore, at a volume of 1000 cm^3 and a pressure of 726 mmHg, the temperature at which the gas would occupy this volume is approximately 429.56 Kelvin.