If a gas at 1.2 atm of pressure and 22.0 degrees of Celsius occupies 0.350L, what pressure will hold the same sample of gas in a volume of 0.050L if the temperature of the gas increases 25 degrees of Celsius?
(P1V1/T1) = (P2V2/T2)
To solve this problem, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in atmospheres)
V = volume (in liters)
n = number of moles of gas
R = ideal gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)
First, let's convert the temperature from degrees Celsius to Kelvin:
T1 = 22.0 °C + 273.15 = 295.15 K (initial temperature)
T2 = 25 °C + 273.15 = 298.15 K (final temperature)
Next, we can rearrange the ideal gas law equation to solve for the new pressure (P2) at the final volume (V2):
(P1 * V1) / (n * T1) = (P2 * V2) / (n * T2)
Since the number of moles of gas (n) and the gas constant (R) are constant, we can simplify the equation:
(P1 * V1) / T1 = (P2 * V2) / T2
Now, let's plug in the given values:
P1 = 1.2 atm
V1 = 0.350 L
T1 = 295.15 K
V2 = 0.050 L
T2 = 298.15 K
(1.2 atm * 0.350 L) / 295.15 K = (P2 * 0.050 L) / 298.15 K
Solving for P2:
P2 = (1.2 atm * 0.350 L * 298.15 K) / (0.050 L * 295.15 K)
P2 = 1.2 atm * 0.350 L * 298.15 K / (0.050 L * 295.15 K)
P2 ≈ 1.82 atm
Therefore, the pressure that will hold the same sample of gas in a volume of 0.050 L if the temperature increases by 25 degrees Celsius is approximately 1.82 atm.
To solve this problem, we can use the combined gas law formula, which states:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Where:
P1 and P2 are the initial and final pressures of the gas
V1 and V2 are the initial and final volumes of the gas
T1 and T2 are the initial and final temperatures of the gas
Given:
P1 = 1.2 atm
V1 = 0.350 L
T1 = 22.0 °C
We want to find P2 when:
V2 = 0.050 L
T2 = 22.0 °C + 25 °C = 47.0 °C
Let's plug in the values into the combined gas law formula and solve for P2:
(P1 * V1) / T1 = (P2 * V2) / T2
(1.2 atm * 0.350 L) / (22.0 °C) = (P2 * 0.050 L) / (47.0 °C)
Now, let's solve for P2:
(0.42 atm) / (22.0 °C) = (P2 * 0.050 L) / (47.0 °C)
Cross-multiply:
(P2 * 0.050 L) = (0.42 atm * 47.0 °C) / (22.0 °C)
(P2 * 0.050 L) = 0.9027 atm
Divide both sides by 0.050 L:
P2 = 0.9027 atm / 0.050 L
P2 = 18.05 atm
Therefore, the pressure required to hold the same sample of gas in a volume of 0.050 L when the temperature increases by 25 °C is approximately 18.05 atm.