91Ó°ÊÓ

Given,

Number of moles n = 0.0100mol and the initial temperature$T_1=27C=300K$.

Now, at constant volume we can use the molar heat capacity at a constant volume$C_v$, and to find the heat added to rise the temperature from to $T_{1}to{T}_2=67C=340K$, we use:

$\begin{array}{r}Q=n{C}_v\mathrm{Δ}T\\ =n{C}_v\left(T_2−T_1\right)\end{array}$

We know that the $C_v$of helium is 12.47 J/mol K, and now substituting the values of$n,C_v,T_2$and$T_1$we get:

$\begin{array}{r}Q=n{C}_v\left(T_2−T_1\right)\\ =0.0100\mathrm{mol}×12.47\mathrm{J}/\mathrm{molK}(340\mathrm{K}−300\mathrm{K})\\ =4.99\mathrm{J}\end{array}$

Therefore, the heat added is 4.99 J

Now, to draw the PV diagram for the process, we have the volume is constant while the temperature increases, therefore according to ideal gas law the pressure will also increase, and the resultant PV diagram is:

Here the pressure is constant, so using the molar heat capacity at constant pressure$C_p$ instead of$C_v$ in the above expression we have:

$Q=n{C}_p\left(T_2-T_1\right)$

Here, the$C_p$ for helium is 20.78 J/molK , substituting the values we get:

$\begin{array}{r}Q=n{C}_p\left(T_2−T_1\right)\\ =0.0100\mathrm{mol}×20.78\mathrm{J}/\mathrm{molK}(340\mathrm{K}−300\mathrm{K})\\ =8.31\mathrm{J}\end{array}$

And the required PV diagram where the pressure is constant and volume increasing is shown below:

The change in internal$∆U$can be calculated by using the first law of thermodynamics, where:

$∆U=Q-W$

However, the volume is kept constant therefore the work done is zero.

$W=0$

Now,

localid="1668312378353" $$\begin{gathered} \mathrm{Δ}{U}_a=Q−W \\ =4.99−0 \\ =4.99\mathrm{J} \end{gathered}$$

In part b the$∆U_b$will also be 4.99 J because the increase of Q in part b in used to increase the work done and not the internal energy.

Hence,$∆U$is same for both part and b that is 4.99J