91Ó°ÊÓ

By using the ideal gas equation, i.e.,$\mathit{p}\mathit{v}\mathbf{=}\mathit{n}\mathit{R}\mathit{T}$find the number of moles of oxygen and the final temperature of the sample.

The formula is as follows:

$pv=nRT$

Here p is pressure, v is volume, T is temperature, R is universal gas constant and n is the number of moles.

Initially, convert temperature in Celsius to Kelvin, i.e.

$$\begin{gathered} T_i=40°\text{C} \\ =273.15+40 \\ =313.15\text{K} \end{gathered}$$

According to ideal gas equation

$pv=nRT$

For finding the number of moles of oxygen present, use the initial condition

$p_i{v}_i=nR{T}_i$

$n=\frac{p_i{v}_i}{R{T}_i}$

By putting in the values, we get

$$\begin{gathered} n=\frac{1.01×{10}^5×1×{10}^{-3}}{8.31×313.15} \\ =3.88×{10}^{-2} mol \end{gathered}$$

Hence, the number of moles of oxygen present is $3.88×{10}^{-2}mol.$

According to the ideal gas equation, find the final temperature of the sample.

$p_f{v}_f=nR{T}_f$

By rearranging the given equation for final temperature, we have

$$\begin{gathered} T_f=\frac{p_f{v}_f}{nR} \\ =\frac{1.06×{10}^5×1.5×{10}^{-3}}{3.88×{10}^{-2}×8.31} \\ =\frac{1.59×{10}^2}{32.24×{10}^{-2}} \\ =0.0493×{10}^4 \end{gathered}$$

Solving further, we have

$T_f=493K$

Hence, the temperature of the sample is 493 K.