Question

FIGURE $P20.50$shows the thermal energy of$0.14mol$of gas as a function of temperature. What is$C_v$ for this gas?

Short answer

The $C_v$for gas is$C_V=29\mathrm{J}/\mathrm{mol}·\mathrm{K}$

Step-by-step solution

Definition of thermal energy

Thermal energy is defined as the energy stored within a system that is relevant to its temperature. Heat is the term used to describe the passage of thermal energy.

Thermodynamics is a branch of physics that studies how heat is transferred across systems and how work is performed in the process.

Step 2:Expalation

If the temperature changes by $∆T$, then the thermal energy for a gas changes by equation $(20.30)$in the form

$\Delta E_{\mathrm{th}}=n{C}_V\Delta T$---------(1)

Solve equation $\left(1\right)$for $C_v$to be in the form

$C_V=\frac{\Delta E_{\mathrm{th}}}{n\Delta T}$------------(2)

From figure $P20.50$, the temperature changes from $100°C$to $200°C$. So, the change in temperature is

$\Delta T=200°\mathrm{C}-100°\mathrm{C}=100°\mathrm{C}$

At this change in temperature, the thermal energy changes by

$\Delta E_{\mathrm{th}}=E_f-E_i$

role="math" localid="1648440626213" $=1892J-1492J$

$=400J$

Now, we plug the values for$∆E_{th}$, $n$and $∆T$into equation $\left(2\right)$to get $C_v$by

$C_V=\frac{\Delta E_{\mathrm{th}}}{n\Delta T}$

$=\frac{400\mathrm{J}}{(0.14\mathrm{mol})(100\mathrm{K})}$

$=29\mathrm{J}/\mathrm{mol}·\mathrm{K}$

The thermal energy of gas

The thermal energy of gas is$C_v=29J/mol.K$