Question

A rigid container holds $2.0mol$of gas at a pressure of $1.0atm$and a temperature of $30°C$.

a. What is the container’s volume?

b. What is the pressure if the temperature is raised to $130°C$?

Short answer

a. The container’s volume is $\mathbf{0}\mathbf{.}\mathbf{050}\mathbf{}{\mathit{m}}^{\mathbf{3}}$.

b. The pressure if the temperature is raised to$130°C\mathrm{is}\mathbf{1}\mathbf{.}\mathbf{33}\mathbf{}\mathbf{atm}$

Step-by-step solution

Defining the Ideal Gas Law

Theideal gas law shows the relationship between the four state variables for an ideal gas, namely the volume $V$of the container, the pressure $p$exerted on the gas, the temperature $T$of the gas, and the number of moles $n$of the gas in the container.

The ideal gas law is given by $PV=nRT$
Where $R$is the universal gas constant and in units of SI its value is $8.31J/mol.K$
Solve this for $V$we get : $V=\frac{nRT}{P}$

Temperature must be expressed in Kelvin and pressure in Pascals, so we must convert the units from Celsius. The conversion between the Celsius scale and the Kelvin scale is given by $T_K=T_C+273$

Finding the container’s volume .

(a) Substitute the value for$T_C=30°C$ to get
$$\begin{gathered} T_C=T_C+273 \\ =30°C+273 \\ =303K \end{gathered}$$
To calculate the pressure in $Pa$, we use the following formula $$\begin{gathered} p=(1atm)\frac{1.01325x{10}^{5}Pa}{1atm}5Pa) \\ =1.01325x{10}^{5}Pa \end{gathered}$$
Now we substitute the values for $n,R,T\mathrm{and}p$into the equation$V=\frac{nRT}{P}$ to get:

$$\begin{gathered} V=\frac{nRT}{P} \\ =\frac{(2mol)(8.314J/mol-K)(303K)}{1.01325\times {10}^{5}Pa} \\ =\mathbf{}\mathbf{0}\mathbf{.}\mathbf{050}\mathbf{}{\mathit{m}}^{\mathbf{3}} \end{gathered}$$

Finding the pressure if the temperature is raised to 130°C

(b) As can be seen from the ideal gas law, pressure $p$is directly proportional to temperature$T$. Thus, if the volume is constant for the same gas, we can obtain the relationship for two points in time as follows.
$$\begin{gathered} \frac{p_2}{p_1}=\frac{T_2}{T_1} \\ \Rightarrow p_2=\left(\frac{T_2}{T_1}\right)p_1 \end{gathered}$$
With $T₂=130°+273=403K,T₁=303K,\mathrm{and}p₁=1atm$. It follows that :
$$\begin{gathered} p_2=(\frac{403K}{303K})(1atm) \\ =\mathbf{1}\mathbf{.}\mathbf{33}\mathbf{}\mathit{a}\mathit{t}\mathit{m}\mathbf{} \end{gathered}$$