$n$moles of a diatomic gas with $C_v=\frac{5}{2}R$has initial pressure $p_1$and volume $V_i$. The gas undergoes a process in which the pressure is directly proportional to the volume until the $rms$speed of the molecules has doubled.

a. Show this process on a $pV$diagram.

b. How much heat does this process require? Give your answer in terms of $n$, $p_1$, and $v_i$.

a. The pressure $p$is proportional to the volume $V$so

$p=\alpha V$

where $\alpha$could be some manner of stability. Take note of the following: $v_{rms}\alpha \sqrt{T}$so $T\alpha {v_{rms}}^2$. When $v_{rms}$doubles the Temperature will quadruple i.e. the final temperature will be $T_f=4T_i$. Still want to utilise this result to calculate the final pressure and volume. This is accomplished in the following way: substitute $p=\alpha V$into the ideal gas equation $pV=nRT$:

$\alpha V^2=nRT$

Therefore

$\alpha V_i^2=nR{T}_i,\alpha V_f^2=nR{T}_f$

When we divide these two equations, we get

$\frac{V_f^2}{V_i^2}=\frac{T_f}{T_i}=4$

Which yields

$\frac{V_f}{V_i}=2\Rightarrow V_f=2V_i$

Because pressure is proportionate to volume, it's as simple as that.

$p_f=2p_i$

This process is shown in the $pV$on the figure below.

b. The change in thermal energy must be discovered.$∆E=E_f-E_i$and the work done $W$in this process. The required heat is then calculated using the first rule of thermodynamics.

$Q=\Delta E+W$

The thermal energy in the initial and in the final state is given by

$E_i=\frac{5}{2}nR{T}_i,E_f=\frac{5}{2}nR{T}_f=\frac{5}{2}nR·4T_i$

When we exploited the knowledge that in this procedure, the temperature quadruples as illustrated in section a. From the ideal gas law, we now have $p_i{V}_i=nR{T}_i$so

$E_i=\frac{5}{2}{p}_i{V}_i,E_f=10p_i{V}_i$

This yields for the change in thermal energy

$\Delta E=E_f-E_i=\frac{15}{2}{p}_i{V}_i$ eq (1)

The area under the line in the diagram represents the work done in this phase. $pV$diagram under the line denoting the process and above the $V$axis. This is the area of the shaded trapezius in the figure below. Therefore

$W=\frac{1}{2}\left(p_i+2p_i\right)\left(2V_i-V_i\right)=\frac{3}{2}{p}_i{V}_i$

Combining (eq-1) and (eq-2) with (1st law) we get

$Q=\frac{15}{2}{p}_i{V}_i+\frac{3}{2}{p}_i{V}_i=9p_i{V}_i$

The velocity$Q=9p_i{V}_i$