We have \(P = 455\,torr\), \(n = 4.4\times10^{-2}mol\), and \(T = 331°C\). We need to find V.
Converting the pressure to Pa:
\[P = 455\,torr \times \frac{101325\,Pa}{760\,torr} \approx 6.085\times10^4\,Pa\]
And converting the temperature from Celsius to Kelvin:
\[T = 331°C + 273.15 = 604.15\,K\]
Now we can use the Ideal Gas Law to find V, rearranging the equation and plugging in the values:
\[V = \frac{nRT}{P}\]
\[V = \frac{(4.4\times10^{-2}\,mol)(8.314\,J/(mol\cdot K))(604.15\,K)}{6.085\times10^4\,Pa}\]
\[V \approx 3.631\times10^{-3}\,L \;or\; 3.631\,mL\]
#Step 4: Calculate "T" for the fourth row#
We are given \(P = 745\,mmHg\), \(V = 11.2\,L\), and \(n = 0.401\,mol\). We need to find T.
Converting the pressure to Pa:
\[P = 745\,mmHg\times\frac{101325\,Pa}{760\,mmHg} \approx 9.978\times10^4\,Pa\]
Using the Ideal Gas Law to find T, rearranging the equation and plugging in the values:
\[T = \frac{PV}{nR}\]
\[T = \frac{(9.978\times10^4\,Pa)(11.2\,L)}{(0.401\,mol)(8.314\,J/(mol\cdot K))}\]
\[T \approx 332.45\,K\]
Now we've completed the whole table, and it can be represented like this:
$$
\begin{array}{|llll|}
\hline
P & V & n & T \\
\hline
7.74 \times 10^3\,\text{Pa} & 12.2\,\text{mL} & 0.00371\,\text{mol} & 25°C \\
\hline
2.1736\times10^5\,\text{Pa} & 43.0\,\text{mL} & 0.421\,\text{mol} & 223\,\text{K} \\
\hline
455\,\text{torr} & 3.631\,\text{mL} & 4.4\times10^{-2}\,\text{mol} & 331°C \\
\hline
745\,\text{mmHg} & 11.2\,\text{L}& 0.401\,\text{mol} & 332.45\,K \\
\hline
\end{array}
$$