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Chapter 13: Problem 62

What is the kinetic energy per unit volume in an ideal gas at (a) \(P=1.00\) atm and (b) \(P=300.0\) atm?

Short Answer

Expert verified
Answer: The kinetic energy per unit volume for an ideal gas at (a) 1.00 atm is 1.701 x 10^5 J/m^3, and at (b) 300.0 atm is 5.103 x 10^7 J/m^3.

Step by step solution

01

Convert pressure to Pascal

First, we will convert the pressure from atmospheres to Pascal. For (a) \(P=1.00\) atm and for (b) \(P=300.0\) atm. Use the conversion factor 1 atm = 101325 Pa: (a) \(P=(1.00)(101325 \textrm{ Pa/atm})=101325 \textrm{ Pa}\) (b) \(P=(300.0)(101325 \textrm{ Pa/atm})=30,397,500 \textrm{ Pa}\)
02

Find n, the number of moles per unit volume

We will find the number of moles per unit volume using the ideal gas law equation, \(PV=nRT\). Rearranging the equation, we get \(n = \frac{PV}{RT}\). Since we are assuming standard temperature, \(T = 273.15\) K. (a) \(n = \frac{101325 \textrm{ Pa}}{(8.314 \textrm{ J/mol K})(273.15 \textrm{ K})} = 44.615 \textrm{ mol/m}^3\) (b) \(n = \frac{30,397,500 \textrm{ Pa}}{(8.314 \textrm{ J/mol K})(273.15 \textrm{ K})} = 13,384.66 \textrm{ mol/m}^3\)
03

Calculate the kinetic energy per unit volume

Now we will use the equation for kinetic energy, \(K = \frac{3}{2}nRT\), to find the kinetic energy per unit volume: (a) \(K = \frac{3}{2}(44.615 \textrm{ mol/m}^3)(8.314 \textrm{ J/mol K})(273.15 \textrm{ K}) = 1.701\times10^5 \textrm{ J/m}^3\) (b) \(K = \frac{3}{2}(13,384.66 \textrm{ mol/m}^3)(8.314 \textrm{ J/mol K})(273.15 \textrm{ K}) = 5.103\times10^7 \textrm{ J/m}^3\) So, the kinetic energy per unit volume in an ideal gas at (a) \(P=1.00\) atm is \(1.701\times10^5 \textrm{ J/m}^3\) and at (b) \(P=300.0\) atm is \(5.103\times10^7 \textrm{ J/m}^3\).

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