91Ó°ÊÓ

(a) The ideal gas law depicts the relationship between the volume of the container, the pressure exerted by the gas, the temperature of the gas, and the number of moles of the gas in the container for an ideal gas.

Equation: $PV=nRT$ is the ideal gas law.

Where $R$is the universal gas constant, which has a value of $R=8.31J/mol$in SI units.

For $T$, we solve the above equation in the form $T=\frac{pV}{nR}$. (1)

data-custom-editor="chemistry" ${\mathrm{P}}_1=1\mathrm{atm}=1.013\mathrm{x}{10}^5\mathrm{Pa}\mathrm{and}{\mathrm{V}}_1=100{\mathrm{cm}}^3$are the initial pressure and volume, respectively.

As a result, we utilize equation (1) to get initial temperature :

$$\begin{gathered} T=\frac{pV}{nR} \\ =\frac{(1.013×{10}^{5}Pa)(100×{10}^{-6}{m}^3)}{(0.0050mol)(8.314J/mol·K)} \\ =\boxed{\mathbf{244}\mathbf{}\mathit{K}} \end{gathered}$$

Given figure :

(b) For $p$to be in the form $\frac{nRT}{V}$, we solve the ideal gas law in equation.

$T=2926K$is the second temperature, and role="math" localid="1650724200234" $V_2=300c{m}^3$is the second volume.

As a result, we utilize the above equation to obtain the second pressure,$p_2$, by :

$$\begin{gathered} p_2=\frac{nR{T}_2}{V_2} \\ =\frac{(0.0050mol)(8.314J/mol·K)(2926K)}{4.05×{10}^{5}Pa} \\ ⇒p_2=(4.05x{10}^{5}Pa)×\frac{1atm}{1.013x{10}^{5}Pa} \\ =\boxed{\mathbf{4}\mathbf{}\mathit{a}\mathit{t}\mathit{m}} \end{gathered}$$

Given figure :

(b) For $V$to be in the form$V=\frac{nRT}{p}$, we solve the ideal gas law in equation.

$T_3=2438K$is the second temperature, and $p_3=2.026×{10}^{5}Pa$is the third pressure.

As a result, we utilize the above equation to obtain the Volume, by :

localid="1662449158099" $$\begin{gathered} V_2=\frac{nR{T}_2}{p_2} \\ =\frac{(0.0050mol)(8.314J/mol·K)(2438K)}{2.026×{10}^{5}Pa} \\ =500×{10}^{-6}{m}^3 \\ =\boxed{\mathbf{500}\mathbf{}\mathit{c}{\mathit{m}}^{\mathbf{3}}} \end{gathered}$$