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Chapter 12: Problem 32

0.10 mol of argon gas is admitted to an evacuated \(50 \mathrm{cm}^{3}\) container at \(20^{\circ} \mathrm{C}\). The gas then undergoes an isothermal expansion to a volume of \(200 \mathrm{cm}^{3}\). a. What is the final pressure of the gas? b. Show the process on a \(p V\) diagram. Include a proper scale on both axes.

Short Answer

Expert verified
a. The final pressure of the gas is 1.205 atm. \n b. The process on the \(pV\) diagram follows a hyperbolic curve from the point (4.82 atm, 0.05L) to (1.205 atm, 0.2L). Both axes should have a proper scale indicating pressure and volume change.

Step by step solution

01

Calculate initial pressure

First, it's important to convert the temperature to Kelvin for it to be compatible with the gas constant 'R' which is 0.0821 L路atm/mol路K. Hence, the temperature in Kelvin is \(20^{\circ} C + 273.15 = 293.15 K\). Secondly, convert the volume from cm^3 to liters as 1 cm^3 = 0.001 L. So, the volume in liters is 50 cm^3 * 0.001 = 0.05L. Using the Ideal Gas Law \(P = \frac{nRT}{V}\) you can find the initial pressure. Substituting the values in, P = [(0.10 mol) * (0.0821 L.atm/mol.K) * (293.15K)]/ 0.05L = 4.82 atm.
02

Calculate final pressure

The final volume after the isothermal expansion is given as 200 cm^3 which is equal to 0.20L. In an isothermal process, \(P1V1 = P2V2\). So, rearrasing for P2, \(P2 = \frac{P1V1}{V2}\) and substituting the values gives you \(P2 =\frac{4.82 atm * 0.05L}{0.20L} = 1.205 atm\). This is the final pressure of the gas after the isothermal expansion.
03

Show the process on a \(pV\) diagram

On the \(pV\) diagram, the x-axis represents the volume and the y-axis represents the pressure. The process of isothermal expansion from volume V1 = 0.05L to volume V2 = 0.20L at a constant temperature follows a hyperbolic path. The starting point (P1, v1) is (4.82atm, 0.05L) and the ending point (P2, V2) is (1.205 atm, 0.20L). The path of expansion should trace a hyperbolic curve showing a decrease in pressure with the increase in volume while maintaining the product (P*V=constant) the same.

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