Question

Calculate the pressure exerted by 22g of carbon dioxide in 0.5 \[d{m^3}\] at 300 K using:
(a) Ideal gas law
(b) Van der Waals equation
[Given: \[a = 360kPa\,d{m^6}\,mo{l^{ - 2}}\] and \[b = 40\dfrac{{c{m^3}}}{{mol}}\] ]
A.(a) 12 atm (b) 11.40 atm
B.(a) 24 atm (b) 21.40 atm
C.(a) 48 atm (b) 42.80 atm
D.(a) 24.63 atm (b) 22.10 atm

Answer 1

Hint: To solve this question, we need to first discuss and understand about both the ideal gas law and Van der Waals equation. Then, on the basis of the formula of these equations, we can substitute the given values to obtain the final answers.

Complete Step-by-Step Answer:
Before we move forward with the solution of the given question, let us first understand some important basic concepts.
1.Ideal gas law: Ideal gas law can be understood as a relationship which is established between the important characteristic properties, viz. pressure, temperature, volume and the number of moles. The ideal gas equation is formed by combining Charles’ Law, Boyle’s Law and Avogadro’s Law. It states that the pressure of any gas is directly proportional to the number of moles of the gas in the given sample, temperature of the gas, and is inversely proportional to the volume of the gas. The constant of proportionality used in this equation is known as the gas constant. Mathematically, the ideal gas law can be represented as:
 \[PV = nRT\]
2.Van der Waals equation: The Van der Waals equation is basically a modification of the ideal gas equation. All these issues were resolved by the Van der Waals equation. Mathematically, it can be represented as:
 \[\left( {P + \dfrac{{a{n^2}}}{{{v^2}}}} \right)(V - nb) = nRT\]
Where a and b are constant values which are unique to each gas.
From the data given and the discussion above, the pressure of the given gas can be calculated as:
a)By Ideal gas equation:
 \[PV = nRT\]
We know that the number of moles are calculated as the ratio of the weight of the sample to its molar mass. Hence, the above equation can be written as:
$\Rightarrow$ \[PV = \dfrac{W}{M}RT\]
$\Rightarrow$ \[P = \dfrac{{WRT}}{{MV}}\]
$\Rightarrow$ \[P = \dfrac{{22 \times 0.0821 \times 300}}{{44 \times 0.5}}\]
$\Rightarrow$ P = 24.63 atm

b)By Van der Waals equation:
 \[(P + \dfrac{{a{n^2}}}{{{v^2}}})(V - nb) = nRT\]
$\Rightarrow$ \[P = \dfrac{{nRT}}{{(V - nb)}} - \dfrac{{a{n^2}}}{{{v^2}}}\]
$\Rightarrow$ \[P = \dfrac{{0.5 \times 0.0821 \times 300}}{{0.5 - (0.5 \times 40 \times {{10}^{ - 3}})}} - \dfrac{{360 \times {{0.5}^2}}}{{101.321 \times {{0.5}^2}}}\]
$\Rightarrow$ P = 25.66 – 3.55
$\Rightarrow$ P = 22.1 atm

Hence, Option D is the correct option

Note: The ideal gas equation is not competent to predict the behaviour of real gases. This was because the ideal gas equation assumed that gases consist of point masses which undergo perfect elastic collisions. This issue was addressed in the Van der Waals equation