A container of volume 2.00 liters contains 2.00 moles of an ideal gas at a pressure of 1.50 atmospheres. What is the average kinetic energy per particle in the gas?
A) $3.78 \times 10-22 \mathrm{~J}$
B) $3.96 \times 10-22 \mathrm{~J}$
C) $1.98 \times 10^{-22} \mathrm{~J}$
D) $2.83 \times 10-22 \mathrm{~J}$
E) $2.52 \times 10-22 \mathrm{~J}$

Question

Quizplus · Accepted Answer

The average kinetic energy per particle in an ideal gas is given by the equation:$$\frac{1}{2}mv^2=\frac{3}{2}kT$$where m is the mass of the particle, v is its velocity, k is the Boltzmann constant, and T is the temperature.We can use the ideal gas law to find the temperature of the gas:$$PV=nRT$$where P is the pressure, V is the volume, n is the number of moles, and R is the ideal gas constant.Substituting the given values into the ideal gas law, we get:$$(1.50\text{ atm})(2.00\text{ L})=(2.00\text{ mol})(0.0821\text{ L atm/mol K})(T)$$$$T=116\text{ K}$$Now we can substitute the given values into the equation for the average kinetic energy per particle:$$\frac{1}{2}mv^2=\frac{3}{2}(1.38\times10^{-23}\text{ J/K})(116\text{ K})$$$$\frac{1}{2}mv^2=2.52\times10^{-22}\text{ J}$$Therefore, the average kinetic energy per particle in the gas is $2.52 \times 10-22 \mathrm{~J}$.

A Container of Volume 2 3.78×10−22 J3.78 \times 10-22 \mathrm{~J}3.78×10−22 J B) 3.96×10−22 J3.96 \times 10-22 \mathrm{~J}3.96×10−22 J

A Container of Volume 2 $3.78 \times 10-22 \mathrm{~J}$ B) $3.96 \times 10-22 \mathrm{~J}$