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Concept explainers
(a)
Interpretation:
Pressure exerted by 1.00 mol CO2 at 298 K and 15.0 L with help of ideal gas equation and van der Waals equation has to be determined.
Concept Introduction:
The expression for ideal gas equation for one mole of gas is as follows:
PV=RT (1)
Here,
P is pressure of gas.
V is molar volume of gas.
T is absolute temperature of gas.
R is universal gas constant.
The formula to calculate pressure of one mole of gas is as follows:
P=RT(V−b)−aV2 (2)
Here,
P is pressure of gas.
V is volume of gas.
T is absolute temperature of gas.
R is universal gas constant.
a and b are van der Waals parameters.
(a)
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Explanation of Solution
Rearrange equation (1) to calculate P.
P=RTV (3)
Substitute 15.0 L for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).
P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)15.0 L=1.63 atm
Substitute 15.0 L for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).
P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(15.0 L−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2(15.0 L)2)=1.62 atm
Pressure exerted by 1.00 mol CO2 at 298 K and 15.0 L with help of ideal gas equation and van der Waals equation is 1.63 atm and 1.62 atm respectively.
(b)
Interpretation:
Pressure exerted by 1.00 mol CO2 at 298 K and 0.500 L with help of ideal gas equation and van der Waals equation has to be determined.
Concept Introduction:
Refer to part (a).
(b)
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Explanation of Solution
Rearrange equation (1) to calculate P.
P=RTV (3)
Substitute 0.500 L for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).
P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)0.500 L=48.9 atm
Substitute 0.500 L for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).
P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(0.500 L−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2(0.500 L)2)=38.9 atm≈39 atm
Pressure exerted by 1.00 mol CO2 at 298 K and 0.500 L with help of ideal gas equation and van der Waals equation is 48.9 atm and 39 atm respectively.
(c)
Interpretation:
Pressure exerted by 1.00 mol CO2 at 298 K and 50 mL with help of ideal gas equation and van der Waals equation has to be determined. Also, reliability of ideal gas with pressure has to be determined.
Concept Introduction:
Refer to part (a).
(c)
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Explanation of Solution
Rearrange equation (1) to calculate P.
P=RTV (3)
Substitute 50 mL for V, 298 K for T, and 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R in equation (3).
P=(8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)(50 mL)(10−3 L1 mL)=489 atm
Substitute 50 mL for V, 298 K for T, 8.20574×10−2 L⋅atm⋅K−1⋅mol−1 for R, 3.658 bar⋅L2⋅mol−2 for a and 4.29×10−2 L⋅mol−1 for b in equation (2).
P=((8.20574×10−2 L⋅atm⋅K−1⋅mol−1)(298 K)((50 mL)(10−3 L1 mL)−4.29×10−2 L⋅mol−1))−(3.658 bar⋅L2⋅mol−2((50 mL)(10−3 L1 mL))2)=1.98×103 atm≈2×103 atm
Ideal gas equation and van der Waals equation have same values at low pressures but high differences arise when pressure is high.
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