Question

$1.0mol$of a monatomic gas interacts thermally with $1.0mol$ of an elemental solid. The gas temperature decreases by ${50}^{o}C$ at constant volume. What is the temperature change of the solid?

Short answer

The temperature change the solid in$\mathrm{\Delta }{T}_{\text{solid}}={25}^{\circ }\mathrm{C}$

Step-by-step solution

Step :1 Introduction 

In the microsystem, the potential energy between bonds is zero, 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

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

Where $n$is the number of moles and $R$is the universal gas constant.

Step :2 Explanation 

If the temperature changes by $∆T$, then the thermal energy for solid changes by equation $(20.34)$in the form

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

When the gas interacts thermally with the solid, they reach the thermal equilibrium after time, so we can get an expression for $∆T_{soli{d}_{}}$when the number of moles is the same

$\mathrm{\Delta }{E}_{{\mathrm{th}}_{\mathrm{gas}}}=\mathrm{\Delta }{E}_{{\mathrm{th}}_{\text{solid}}}$

$\frac{3}{2}nR\mathrm{\Delta }{T}_{\text{gas}}=3nR\mathrm{\Delta }{T}_{\text{solid}}$

$\mathrm{\Delta }{T}_{\text{solid}}=\frac{\mathrm{\Delta }{T}_{\text{gas}}}{2}$

Step : 3 Values in equation 

Now, we plug the values for $∆T_{gas}$into equation (3) to get $∆T_{solid}$

$\mathrm{\Delta }{T}_{\text{solid}}=\frac{\mathrm{\Delta }{T}_{\text{gas}}}{2}$

$=\frac{{50}^{\circ }\mathrm{C}}{2}$

$={25}^{\circ }\mathrm{C}$