Container in a Fireplace!

Use the Ideal Gas Law to relate the final pressure to the temperature and the initial pressure.

The Ideal Gas Law can be written as:

\begin{equation*}
PV=nRT,
\end{equation*}

where for two states can be written as:

\begin{equation*}
\frac{n_i R T_i}{P_i}=\frac{n_f R T_f}{P_f}.
\end{equation*}

Solving for \(P_f\) we get:

\begin{equation*}
P_f=P_i \frac{T_f}{T_i},
\end{equation*}

which, with numerical values, is:

\begin{equation*}
P_f \approx 2.7\,\text{atm}.
\end{equation*}

For a more detailed explanation of any of these steps, click on “Detailed Solution”.

We need to calculate the pressure inside the container after it reaches a temperature of 127\(^\circ\)C. In order to solve this, we must first identify the thermodynamic process between the initial condition and the final condition. Because the problem already told us that any change in volume is neglected, then we are dealing with a constant volume process. To obtain the answer to this problem we must use the ideal gas law

\begin{equation}
\label{gasideal}
PV=nRT,
\end{equation}

where \(P\) is the absolute pressure inside the container, \(V\) its volume, \(n\) the number of moles of gas inside the container, \(R\) the ideal gas constant and \(T\) the gas absolute temperature.

We can now write equation \eqref{gasideal}, taking into account the initial condition (container at room temperature), for which we will use the subscript \(i\) as follows

\begin{equation}
\label{gasi}
P_iV_i=n_iRT_i.
\end{equation}

We can also write a similar expression for the final condition (container in the fireplace), where we will use the subscript \(f\), so

\begin{equation}
\label{gasf}
P_fV_f=n_fRT_f.
\end{equation}

From the prompt, we know that the volume between the initial and final condition is unchanged, so we can express this as

\begin{equation}
\label{igualdad}
V_i=V_f.
\end{equation}

From equation \eqref{gasi}, we can solve for \(V_i\) to get

\begin{equation}
\label{vi}
V_i=\frac{n_iRT_i}{P_i}.
\end{equation}

The same can be done to solve for \(V_f\) from equation \eqref{gasf}; explicitly

\begin{equation}
\label{vf}
V_f=\frac{n_f R T_f}{P_f}.
\end{equation}

Substituting the results from equations \eqref{vi} and \eqref{vf} into equation \eqref{igualdad}, we obtain the following expression

\begin{equation}
\label{expf}
\frac{n_i R T_i}{P_i}=\frac{n_f R T_f}{P_f}.
\end{equation}

Because no gas leaves or enters the container during the thermodynamic process, the number of moles in the initial condition and final condition are the same; hence,

\begin{equation}
\label{moles}
n_i=n_f.
\end{equation}

Using the condition given in equation \eqref{moles} to simplify the equation \eqref{expf}, we obtain:

\begin{equation}
\frac{RT_i}{P_i}=\frac{R T_f}{P_f},
\end{equation}

where we can cancel out the constant \(R\) to finally get

\begin{equation}
\label{expf2}
\frac{T_i}{P_i}=\frac{T_f}{P_f},
\end{equation}

an expression which is valid for any ideal gas that undergoes a thermodynamic process at a constant volume. The problem asks about the pressure inside the container when it is on the fireplace, that is \(P_f\). Solving from equation \eqref{expf2} for \(P_f\), we have that

\begin{equation}
P_f\frac{T_i}{P_i}=T_f,
\end{equation}

\begin{equation}
\label{expf3}
P_f=P_i \frac{T_f}{T_i}.
\end{equation}

Now, we can use the numerical values given in the problem to calculate explicitly \(P_f\). Remember that \(T_i\) and \(T_f\) are absolute temperatures, so they must be given in Kelvin. Calculating each temperature by using the conversion between Celsius and Kelvin, we have

\begin{equation}
\label{t1}
T_1=(23+273.15)\,\text{K}=296.15\,\text{K},
\end{equation}

and

\begin{equation}
\label{t2}
T_2=(127+273.15)\,\text{K}=400.15\,\text{K}.
\end{equation}

Using the numerical values from equations \eqref{t1} and \eqref{t2} into equation \eqref{expf3}, we obtain

\begin{equation}
P_f=(2\,\text{atm})\frac{400.15\,\text{K}}{296.15\,\text{K}}\approx 2.7\,\text{atm}.
\end{equation}

Thus, the pressure increased by approximately \(0.7\,\text{atm}\).

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