91Ó°ÊÓ

Here, the work is done by the gas on the piston, hence the work done is positive. This work at constant pressure could be expressed by the:

$W=p\left(V_2-V_1\right)$………………. (i)

$W=P∆V$…………………. (ii)

Given,

Moles n = 6

Initial Temperature$T_1=27C°$

To calculate temperature T , we will use the ideal gas law.

$PV=nRT$

$V=\frac{nRT}{P}$………………. (iii)

Here, P is pressure, V is volume, n is the number of moles, R is the gas constant or 8.314J/mol.K

Now from equation ii we have seen that when the in temperature and volume change the formula will become

$∆V=\frac{nR∆T}{p}$

here,$\frac{nR}{T}$is constant, hence we can substitute the value$∆V$in equation (i).

Now, equation (i) becomes:$W=p∆V$

$$\begin{gathered} W=\frac{p×nR∆T}{p} \\ W=nR\left[T_2-T_1\right] \end{gathered}$$

now by substituting the values of n,R,$T_1$ is equation (iii), for $T_2$we get.

$$\begin{gathered} T_2=\frac{T_1+W}{nR} \\ =27C°+2.4*{10}^{3}J/mol.K \\ =75.1C° \end{gathered}$$

Therefore, the final temperature of the gas is$75.1C°$