An ideal gas at 18.5°C and a pressure of 1.54 105 Pa occupies a volume of 2.10 m3.
(a) How many moles of gas are present?
moles
(b) If the volume is raised to 5.30 m3 and the temperature raised to 31.0°C, what will be the pressure of the gas?
Pa
I used the formula pv = nRT
(1.54*10^5)(2.10m^3 = n(8.31)(18.5)
I did the the algebra and got 2103 moles, but that doesn't seem to be the answer. What am I doing wrong?
Temperature has to be in Kelvins,
To solve this problem using the ideal gas equation (PV = nRT), you are on the right track. However, it seems that there might be a small calculation error in your solution. Let's break down each part of the question to find the correct answer.
(a) How many moles of gas are present?
To find the number of moles (n), we can rearrange the ideal gas equation to solve for n:
n = PV / (RT)
Plugging in the given values:
P = 1.54 × 10^5 Pa
V = 2.10 m^3
R = 8.31 J/(mol·K)
T = 18.5 °C = 18.5 + 273.15 K (convert to Kelvin)
n = (1.54 × 10^5 Pa) × (2.10 m^3) / (8.31 J/(mol·K) × (18.5 + 273.15) K)
Performing the calculation:
n ≈ 0.1391 mol
Therefore, there are approximately 0.1391 moles of gas present.
For part (b), we can use the same formula and solve for the new pressure:
P1V1 / T1 = P2V2 / T2
Plugging in the given values:
P1 = 1.54 × 10^5 Pa (initial pressure)
V1 = 2.10 m^3 (initial volume)
T1 = 18.5 + 273.15 K (initial temperature in Kelvin)
P2 = ? (final pressure, what we are trying to find)
V2 = 5.30 m^3 (final volume)
T2 = 31.0 + 273.15 K (final temperature in Kelvin)
Now, we can rearrange the equation to solve for P2:
P2 = (P1 × V1 × T2) / (V2 × T1)
Plugging in the given values and performing the calculation:
P2 ≈ (1.54 × 10^5 Pa × 2.10 m^3 × (31.0 + 273.15) K) / (5.30 m^3 × (18.5 + 273.15) K)
P2 ≈ 1.07 × 10^5 Pa
Therefore, the pressure of the gas when the volume is 5.30 m^3 and the temperature is 31.0°C is approximately 1.07 × 10^5 Pa.