91Ó°ÊÓ

Change in Temperature$\left(\mathrm{Δ}T\right)$

of gas.${10}^{0}c$

formula used: The change of thermal energy$\left(\mathrm{Δ}{E}_{\text{th}}\right)$can be defined as:

$\mathrm{Δ}{E}_{th}=mc\mathrm{Δ}T=nC\mathrm{Δ}T$

For the case of gases, consider the equation$\left(1\right)$as follows:

$\mathrm{Δ}{E}_{th}=nC\mathrm{Δ}T$

This means that more thermal energy$\left(\mathrm{Δ}{E}_{th}\right)$is required to change the temperature $\left(\mathrm{Δ}T\right)$when the molar specific heat $C$is high, whereas for small molar specific heat, $C$the thermal energy$\left(\mathrm{Δ}{E}_{\text{th}}\right)$required is lower. For a molar specific heat at a constant pressure$C_P$, the change of thermal energy $\left(\mathrm{Δ}{E}_{thp}\right)$we have:

$\mathrm{Δ}{E}_{th,p}=n{C}_P\mathrm{Δ}T$

$\left(3\right)$

Similarly, the change of thermal energy $\left(\mathrm{Δ}{E}_{th,V}\right)$for a process with constant volume is:

$\mathrm{Δ}{E}_{th,V}=n{C}_V\mathrm{Δ}T$

$\left(4\right)$

But:

$C_P>C_V$

$\left(5\right)$

Now, isolating $\left(C_P\right)$from equation$\left(3\right)$and $\left(C_V\right)$from equation $\left(4\right)$we have:

$C_P=\frac{\mathrm{Δ}{E}_{op}P}{n\mathrm{Δ}T}$

(6)

And

$C_V=\frac{\mathrm{Δ}{E}_{ith}V}{n\mathrm{Δ}T}$

(7)

Replacing $\left(6\right)$and $\left(7\right)$into equation$\left(5\right)$we obtain:

$\mathrm{Δ}{E}_{th,P}>\mathrm{Δ}{E}_{th,V}$

$\left(8\right)$

The gas must be heated to a constant volume in order to consume the least amount of thermal energy.

According to result $\left(8\right)$to use the least amount of thermal energy, the gas must be heated to constant volume.

$W=\frac{(0.10\mathrm{mol})\left(8.314\frac{\mathrm{J}}{\mathrm{Kmol}}\right)(573\mathrm{K})\left(0.001{\mathrm{m}}^3\right)}{0.002{\mathrm{m}}^3}W=238.15\mathrm{J}$

It's an easy computation; all you have to worry about is the unit system.

$$\begin{gathered} V_i=0.002{\mathrm{m}}^3 \\ {V}_f=0.001{\mathrm{m}}^3 \\ n=0.10\mathrm{mol} \\ {T}_i=473\mathrm{K} \\ R=8.314\frac{\mathrm{J}}{\mathrm{kmol}} \end{gathered}$$

b) We need to determine the expression for the work on the gas at constant temperature, and we know that the differential work on a gas is provided by:

$dW=PdV$

As the gas is ideal we can put the pressure in terms of volume and temperature:

$P=\frac{nR{T}_i}{V}$

Then the differential work is given by:

$dW=\frac{nR{T}_{i}dV}{V}$

By integrating the expression, we can calculate the work on the gas:

$dW=∫\frac{nR{T}_{i}dV}{V}$

Then the work is given by:

$W-nR{T}_i\ln \left(\frac{V_f}{V_i}\right)$

$$\begin{gathered} W=(0.10\mathrm{mol})\left(8.314\frac{J}{kmol}\right)(573\mathrm{K})\ln \left(\frac{0.001{\mathrm{m}}^3}{0.002{\mathrm{m}}^3}\right) \\ W=330.21J \end{gathered}$$

It's a simple calculation; the only thing to consider is the unit system.