91Ó°ÊÓ

The processes that take place in a perfectly insulated chamber are called adiabatic processes. This means that in the adiabatic process, heat can neither enter the chamber nor can it leave. So, all the work done on the system is stored in the system in the form of internal energy.

1. No. of moles$\text{n}=2.0\text{″¾´Ç±ô}$
2. Temperature, $\text{T}=15.0\text{‿é}$

The internal energy change is given as-

$\mathrm{Δ}{E}_{\mathrm{int}}=n{C}_v\mathrm{Δ}T$

Here$C_v$Is molar-specific heat at constant volume.

Work done is given by-

$W=Q−\mathrm{Δ}{E}_{\mathrm{int}}$

Here Q is heat transferred

For an adiabatic process, heat transfer is zero. Hence,

$$\begin{gathered} Q=0 \\ \begin{array}{c}W=−\mathrm{Δ}{E}_{\mathrm{int}}\\ =−n{C}_v\mathrm{Δ}T\end{array} \end{gathered}$$

The gas in the problem is an ideal monoatomic gas, so for a monoatomic ideal gas, themolar-specific heat is

$C_v=\frac{3}{2}R$

And hence, the equation becomes

$\begin{array}{c}W=−\mathrm{Δ}{E}_{\mathrm{int}}\\ =−n{C}_v\mathrm{Δ}T\\ =−\frac{3}{2}nR\mathrm{Δ}T\end{array}$

So, work done is

$\begin{array}{c}W=−\frac{3}{2}\left(2.0\text{″¾´Ç±ô}\right)\left(8.31\text{ }\frac{\text{J}}{\text{mol}}\text{.K}\right)\left(15.0\text{‿é}\right)\\ =−373.95\text{ J}\\ â‰\dots −374\text{ J}\end{array}$

$\text{Q}=0.$

$\begin{array}{c}\mathrm{Δ}{E}_{\mathrm{int}}=n{C}_v\mathrm{Δ}T\\ =\frac{3}{2}nR\mathrm{Δ}T\\ =374\text{ J}\end{array}$

$\begin{array}{c}\mathrm{Δ}K=\frac{\mathrm{Δ}{E}_{\mathrm{int}}}{N}\\ =\frac{\mathrm{Δ}{E}_{\mathrm{int}}}{n{N}_A}\end{array}$

$N_A$is Avogadro’s number.

$\mathrm{Δ}K=\frac{374\text{ J}}{\left(2.0\text{″¾´Ç±ô}\right)(6.02×{10}^{23}{\text{″¾´Ç±ô}}^{−1})}=3.11×{10}^{−22}\text{ J}$

The work done, the heat transferred, the change in internal energy of the gas and the average kinetic energy per atom of the gas, are $−374\text{ J,0 J,}374\text{ J,²¹²Ô»å}3.11×{10}^{−22}\text{ J}$ respectively.