Oxygen gas having a volume of$\mathbf{1000}\mathbf{}\mathit{c}{\mathit{m}}^{\mathbf{3}}$at$\mathbf{40}\mathbf{°}\mathit{C}$and$\mathbf{1}\mathbf{.}\mathbf{01}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathit{P}\mathit{a}$ expands until its volume is$\mathbf{1500}\mathbf{}\mathit{c}{\mathit{m}}^{\mathbf{3}}$and its pressure is$\mathbf{1}\mathbf{.}\mathbf{06}\mathbf{\times }{\mathbf{10}}^{\mathbf{5}}\mathbf{}\mathit{P}\mathit{a}$. Find

(a) The number of moles of oxygen present

(b) Final temperature of the sample.

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

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

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

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

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

Solving further, we have

$T_f=493K$

Hence, the temperature of the sample is 493 K.