The thermal energy of $1.0Mol$ of a substance is increased by $1.0J$. What is the temperature change if the system is (a) a monatomic gas, (b) a diatomic gas, and (c) a solid?

(a) The potential energy between the bonds is zero in the microsystem and the kinetic energy of a monatomic gas is translational kinetic energy. If the temperature changes by $\mathrm{\Delta }T$, then the thermal energy for monatomic gas changes by equation $(20.34)$in the form

$\mathrm{\Delta }{E}_{\mathrm{th}}=\frac{3}{2}nR\mathrm{\Delta }T$

Where $n$is the number of moles and $R$is the universal gas constant. We solve equation (1) for $\mathrm{\Delta }T$ to be in the form

$\mathrm{\Delta }T=\frac{2}{3}\frac{\mathrm{\Delta }{E}_{\text{th}}}{nR}$

Now, plug the values for $n,\mathrm{\Delta }{E}_{\text{th}}$and $R$into equation (2) to get $\mathrm{\Delta }T$

$\mathrm{\Delta }T=\frac{2}{3}\frac{\mathrm{\Delta }{E}_{\mathrm{th}}}{nR}$

$=\frac{2}{3}\left[\frac{1.0\mathrm{J}}{\left(1.0\mathrm{mol}\right)\left(8.314\mathrm{J}/{\mathrm{mol}}^{-1}\cdot {\mathrm{K}}^{-1}\right)}\right]$

$=0.080K$

(b) If the temperature changes by$\mathrm{\Delta }T$, then the thermal energy for diatomic gas changes by equation (20.37) in the form

$\mathrm{\Delta }{E}_{\text{th}}=\frac{5}{2}nR\mathrm{\Delta }T$

We solve equation

(3) for $\mathrm{\Delta }T$to be in the form

$\mathrm{\Delta }T=\frac{2}{5}\frac{\mathrm{\Delta }{E}_{\mathrm{th}}}{nR}$

Now, plug the values for $n,\mathrm{\Delta }{E}_{\mathrm{th}}$and $R$into equation (4) to get$\mathrm{\Delta }T$

$\mathrm{\Delta }T=\frac{2}{5}\frac{\mathrm{\Delta }{E}_{\text{th}}}{nR}$

$=\frac{2}{5}\left[\frac{1.0\mathrm{J}}{\left(1.0\mathrm{mol}\right)\left(8.314\mathrm{J}/{\mathrm{mol}}^{-1}\cdot {\mathrm{K}}^{-1}\right)}\right]$

$=0.048K$

(c) If the temperature changes by$\mathrm{\Delta T}$, then the thermal energy for solid changes by equation (20.34) in the form

$\mathrm{\Delta }{E}_{\mathrm{th}}=3nR\mathrm{\Delta }T$

Solve equation for$\mathrm{\Delta }T$

$\mathrm{\Delta }T=\frac{\mathrm{\Delta }{E}_{\mathrm{th}}}{3nR}$

Now, plug the values for $n,\mathrm{\Delta }{E}_{\mathrm{th}}$and $R$into equation (6) to get$\mathrm{\Delta }T$

$\mathrm{\Delta }T=\frac{\mathrm{\Delta }{E}_{\mathrm{th}}}{3nR}$

$=\left[\frac{1.0\mathrm{J}}{3\left(1.0\mathrm{mol}\right)\left(8.314\mathrm{J}/{\mathrm{mol}}^{-1}\cdot {\mathrm{K}}^{-1}\right)}\right]$

$=0.040K$