Question

An ideal gas at 0 degrees C consists of 1.0*10^23 atoms. 22J of thermal energy are added to the gas.What is the new temperature in degrees C ?

Answer 1

Concepts and reason

The concept of change in internal energy is required to solve the problem.

First, calculate the number of moles by using the number of atoms and Avogadro number. Finally, use the expression of change in internal energy and determine the new temperature of the gas.

Fundamentals

The change in internal energy is given by the relation,

$\Delta U = \frac{3}{2}nR\Delta T$

Here, n is the number of moles, R is the gas constant, and $\Delta T$ is the change in temperature.

The change in temperature is given as,

$\Delta T = {T_2} - {T_1}$

Here, ${T_1}$ is the initial temperature and ${T_2}$ is the final temperature.

The number of moles of a gas is determined by using the relation,

$n = \frac{N}{{{N_{\rm{A}}}}}$

Here, N is the number of atoms and ${N_{\rm{A}}}$ is the Avogadro number.

Calculate the number of moles.

The number of moles of the gas is calculated by using the relation,

$n = \frac{N}{{{N_{\rm{A}}}}}$

Here, N is the number of atoms and ${N_{\rm{A}}}$ is the Avogadro number.

Substitute $1.0 \times {10^{23}}{\rm{ atoms}}$ and $6.023 \times {10^{23}}{\rm{ moles/atoms}}$ in the above equation and solve for n.

$\begin{array}{c}\\n = \frac{{\left( {1.0 \times {{10}^{23}}{\rm{ atoms}}} \right)}}{{\left( {6.023 \times {{10}^{23}}{\rm{ atoms/moles}}} \right)}}\\\\ = 0.166{\rm{ mol}}\\\end{array}$

The change in internal energy of the gas is given as,

$\Delta U = \frac{3}{2}nR\Delta T$

Here, n is the number of moles of the gas and $\Delta T$ is the change in temperature of the gas.

Substitute ${T_2} - {T_1}$ for $\Delta T$ in the above equation $\Delta U = \frac{3}{2}nR\Delta T$ and solve for ${T_2}$ .

$\begin{array}{c}\\\Delta U = \frac{3}{2}nR\left( {{T_2} - {T_1}} \right)\\\\{T_2} - {T_1} = \frac{{2\Delta U}}{{3nR}}\\\\{T_2} = \frac{{2\Delta U}}{{3nR}} + {T_1}\\\end{array}$

Here, ${T_1}$ is the initial temperature of the gas and ${T_2}$ is the new temperature of the gas.

Substitute 22 J for $\Delta U$ , $0{{\rm{ }}^{\rm{o}}}{\rm{C}}$ for ${T_1}$ , 0.166 moles for n, and $8.314{\rm{ J/mol}}{ \cdot ^{\rm{o}}}{\rm{C}}$ for R in the equation ${T_2} = \frac{{2\Delta U}}{{3nR}} + {T_1}$ .

$\begin{array}{c}\\{T_2} = \frac{{2\left( {22\;{\rm{J}}} \right)}}{{3\left( {0.166{\rm{ mol}}} \right)\left( {8.314{\rm{ J/mol}}{ \cdot ^{\rm{o}}}{\rm{C}}} \right)}} + 0{{\rm{ }}^{\rm{o}}}{\rm{C}}\\\\{\rm{ = }}10.6{{\rm{ }}^{\rm{o}}}{\rm{C}}\\\end{array}$

Ans:

The new temperature of the gas is $10.6{{\rm{ }}^{\rm{o}}}{\rm{C}}$ .