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
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To solve this problem, we can use the ideal gas law equation:
PV = nRT
where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature in Kelvin
Now let's solve the problem step by step:
(a) To find the number of moles (n), we need to rearrange the ideal gas law equation as:
n = PV / RT
Given:
P = 1.92 x 10^5 Pa
V = 2.10 m^3
T = 6.0°C = 6.0 + 273.15 = 279.15 K
R is the ideal gas constant and has a value of 8.314 J/(mol·K).
Now substitute the values into the equation:
n = (1.92 x 10^5 Pa) x (2.10 m^3) / (8.314 J/(mol·K) x 279.15 K)
Simplify the equation:
n = 1.92 x 10^5 x 2.10 / (8.314 x 279.15)
Calculate the value:
n ≈ 145.64 moles
Therefore, there are approximately 145.64 moles of gas present.
(b) To find the pressure of the gas (P), we can use the rearranged ideal gas law equation:
P = nRT / V
Given:
V = 4.60 m^3
T = 30.5°C = 30.5 + 273.15 = 303.65 K
Using the value of n from part (a) and the given values, substitute these into the equation:
P = (145.64 moles) x (8.314 J/(mol·K)) x (303.65 K) / (4.60 m^3)
Simplify the equation:
P = (145.64 x 8.314 x 303.65) / 4.60
Calculate the value:
P ≈ 3.21 x 10^5 Pa
Therefore, the pressure of the gas is approximately 3.21 x 10^5 Pa.