A Sample of gas originally at 25 degrees C and 1 atm pressure in a 2.5 L container is allowed to expand until the pressure is .85 atm and the temperature is 15 degrees C. What is the final volume of gas?
Did anyone get 2.6 L
No. I have 2.8.
To find the final volume of the gas, we can use the ideal gas law equation, which states:
PV = nRT
where:
P = pressure of the gas
V = volume of the gas
n = number of moles of the gas
R = ideal gas constant (0.0821 L·atm/mol·K)
T = temperature in Kelvin
First, let's convert the temperatures to Kelvin:
Initial temperature (T1) = 25°C + 273.15 = 298.15 K
Final temperature (T2) = 15°C + 273.15 = 288.15 K
Now, let's calculate the initial number of moles of the gas using the ideal gas law:
P1V1 = nRT1
n = (P1V1) / (R * T1)
n = (1 atm * 2.5 L) / (0.0821 L·atm/mol·K * 298.15 K)
n ≈ 0.101 mol
Next, let's use the same equation to find the final volume:
(P2V2) / (R * T2) = n
(0.85 atm * V2) / (0.0821 L·atm/mol·K * 288.15 K) = 0.101 mol
Now, we solve for V2:
V2 = (0.101 mol * 0.0821 L·atm/mol·K * 288.15 K) / 0.85 atm
V2 ≈ 2.459 L
Therefore, the final volume of gas is approximately 2.459 L, which is similar to the given value of 2.6 L.
To find the final volume of the gas, we can use the combined gas law equation, which relates the initial and final states of a gas:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 = initial pressure (1 atm)
V1 = initial volume (2.5 L)
T1 = initial temperature (25 degrees C + 273.15 = 298.15 K)
P2 = final pressure (0.85 atm)
V2 = final volume (unknown)
T2 = final temperature (15 degrees C + 273.15 = 288.15 K)
Now, let's plug in the values we know into the equation:
(1 atm * 2.5 L) / (298.15 K) = (0.85 atm * V2) / (288.15 K)
To find the final volume (V2), we can rearrange the equation:
(0.85 atm * V2) = (1 atm * 2.5 L * 288.15 K) / (298.15 K)
V2 = (1 atm * 2.5 L * 288.15 K) / (298.15 K * 0.85 atm)
V2 ≈ 2.399 L
Therefore, the final volume of the gas is approximately 2.399 L, not 2.6 L.