Answer:
1.78 atm
Explanation:
From the question given above, the following data were obtained:
Mass of Cl = 355 g
Molecular weight of Cl = 70.9 g/mol
Volume (V) = 70.0 L
Temperature (T) = 30 °C
Pressure (P) =?
Next, we shall determine the number of mole in 355 g of chlorine (Cl). This can be obtained as follow:
Mass of Cl = 355 g
Molecular weight of Cl = 70.9 g/mol
Mole of Cl =.?
Mole = mass /molecular weight
Mole of Cl = 355 /70.9
Mole of Cl = 5 moles
Next, we shall convert 30 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
Temperature (T) = 30 °C
Temperature (T) = 30 °C + 273 = 303 K
Finally, we shall determine the pressure of the gas as illustrated below:
Volume (V) = 70.0 L
Number of mole (n) = 5 moles
Temperature (T) = 303 K
Gas constant (R) = 0.0821 atm.L/Kmol
Pressure (P) =?
PV = nRT
P × 70 = 5 × 0.0821 × 303
P × 70 = 124.3815
Divide both side by 70
P = 124.3815 / 70
P = 1.78 atm
Therefore, the pressure of the gas is 1.78 atm
The pressure of the gas in atmospheres if it's temperature is 30.0 degrees celcius is 1.77 atm.
Pressure of the given gas will be calculated by using the ideal gas equation as:
PV = nRT, where
P = pressure of chlorine gas = ?
V = volume of chlorine gas = 70 L
R = universal gas constant = 0.082 L.atm / K.mol
T = temperature chlorine = 30 degree celsius = 303 K
n is the moles of gas which can be calculated as:
n = W/M, where
W = given mass of chlorine gas = 355g
M = molar mass of chlorine gas = 70.9g/mol
n = 355/70.9 = 5 moles
On putting all these values on the above equation and calculate for P, we get
P = (5)(0.082)(303) / 70
P = 1.77 atm
Hence required pressure is 1.77 atm.
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