An ideal gas at 6.0°C and a pressure of 1.92 105 Pa occupies a volume of 2.10 m3.
(a) How many moles of gas are present?
moles
(b) If the volume is raised to 4.60 m3 and the temperature raised to 30.5°C, what will be the pressure of the gas?
Pa
I was able to get 173.81 for the mole present but I'm not able to get part (B)
To solve part (b), we can use the ideal gas law equation:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
First, let's convert the given temperature of 6.0°C to Kelvin:
T1 = 6.0°C + 273.15 = 279.15 K
Next, we can calculate the initial number of moles using the ideal gas law:
n1 = (PV1) / (RT1)
= (1.92 * 10^5 Pa * 2.10 m^3) / ((8.314 J/(mol*K)) * 279.15 K)
Now, let's convert the new temperature of 30.5°C to Kelvin:
T2 = 30.5°C + 273.15 = 303.65 K
We can rearrange the ideal gas law equation to solve for the new pressure (P2):
P2 = (n1 * R * T2) / V2
Substituting the values, we get:
P2 = (173.81 mol * 8.314 J/(mol*K) * 303.65 K) / 4.60 m^3
Calculating this expression will give us the answer to part (b).
To determine the number of moles of gas present, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in pascals)
V = volume (in cubic meters)
n = number of moles of gas
R = ideal gas constant (8.314 J/(mol·K))
T = temperature (in Kelvin)
We need to convert the temperature in Celsius to Kelvin by adding 273.15:
T1 = 6°C + 273.15 = 279.15 K
Now we can calculate the number of moles of gas using the given information:
(1.92×10^5 Pa)(2.10 m^3) = n(8.314 J/(mol·K))(279.15 K)
Solving for n:
(1.92×10^5 Pa)(2.10 m^3) = n(8.314 J/(mol·K))(279.15 K)
n = (1.92×10^5 Pa)(2.10 m^3) / (8.314 J/(mol·K))(279.15 K)
n ≈ 173.81 moles
For part (b), we can use the combined gas law equation to find the new pressure of the gas when the volume is raised to 4.60 m^3 and the temperature is raised to 30.5°C:
(P1)(V1) / (T1) = (P2)(V2) / (T2)
Again, we need to convert the temperature in Celsius to Kelvin:
T1 = 6°C + 273.15 = 279.15 K
T2 = 30.5°C + 273.15 = 303.65 K
Now we can plug in the values and solve for the pressure (P2):
(1.92×10^5 Pa)(2.10 m^3) / (279.15 K) = (P2)(4.60 m^3) / (303.65 K)
P2 = [(1.92×10^5 Pa)(2.10 m^3) / (279.15 K)] * (303.65 K / 4.60 m^3)
P2 ≈ 3.27×10^5 Pa
Therefore, the pressure of the gas is approximately 3.27×10^5 Pa.