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Example 1.2 In an industrial process, nitrogen is heated to 500 ..., Summaries of Industrial Chemistry

volume of CO2 at 500 K and 100 atm by treating it as a van der Waals gas. Answer. The van der Waals equation is. Vm = V/n as ...

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Example 1.2
In an industrial process, nitrogen is heated to 500 K in a vessel of constant volume. If it enters
the vessel at 100 atm and 300 K, what pressure would it exert at the working temperature if it
behaved as a perfect gas?
Solution
Example 1.3
The mass percentage composition of dry air at sea level is approximately N2: 75.5; O2: 23.2;
Ar: 1.3. What is the partial pressure of each component when the total pressure is 1.20 atm?
Answer
Example 1.4 Using the van der Waals equation to estimate a molar volume Estimate the molar
volume of CO2 at 500 K and 100 atm by treating it as a van der Waals gas.
Answer
The van der Waals equation is
Vm = V/n as
pf3
pf4
pf5

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Example 1. In an industrial process, nitrogen is heated to 500 K in a vessel of constant volume. If it enters the vessel at 100 atm and 300 K, what pressure would it exert at the working temperature if it behaved as a perfect gas? Solution

Example 1. The mass percentage composition of dry air at sea level is approximately N2: 75.5; O2: 23.2; Ar: 1.3. What is the partial pressure of each component when the total pressure is 1.20 atm? Answer

Example 1.4 Using the van der Waals equation to estimate a molar volume Estimate the molar volume of CO2 at 500 K and 100 atm by treating it as a van der Waals gas. Answer The van der Waals equation is

Vm = V/n as

According to Table 1.6, a = 3.610 dm^6 atm mol−2^ and b = 4.29 × 10−2^ dm^3 mol−1. Under the stated conditions, RT/p = 0.410 dm^3 mol−1. The coefficients in the equation for Vm are therefore b + RT/p = 0.453 dm^3 mol− a/p = 3.61 × 10−2^ (dm^3 mol−1)^2 ab/p = 1.55 × 10−3 (dm^3 mol−1)^3 Therefore, on writing x = Vm/( dm^3 mol−1), the equation to solve is x^3 − 0.453x^2 + (3.61 × 10−2)x − (1.55 × 10−3) = 0 The acceptable root is x = 0.366 (Fig. 1.18), which implies that Vm = 0.366 dm^3 mol−1. For a perfect gas under these conditions, the molar volume is 0.410 dm^3 mol−1.

1.15(b) A gas at 350 K and 12 atm has a molar volume 12 per cent larger than that calculated from the perfect gas law. Calculate (a) the compression factor under these conditions and (b) the molar volume of the gas. Which are dominating in the sample, the attractive or the repulsive forces?

1.16(b) Cylinders of compressed gas are typically filled to a pressure of 200 bar. For oxygen, what would be the molar volume at this pressure and 25°C based on (a) the perfect gas equation, (b) the van der Waals equation. For oxygen, a = 1.364 dm6 atm mol−2, b = 3.19 × 10−2 dm3 mol−1.

1.22(b) A certain gas obeys the van der Waals equation with a = 0.76 m6 Pa mol−2. Its volume is found to be 4.00 × 10−4^ m3 mol−1^ at 288 K and 4.0 MPa. From this information calculate the van der Waals constant b. What is the compression factor for this gas at the prevailing temperature and pressure?