Question

Iodine, I₂, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. What is the temperature in a 73.3mL bulb that contains 0.292 g of I₂,vapour at a pressure of 0.462 atm?

Short answer

The temperature is \({\rm{(T) = 375}}{\rm{.55\;K}}\).

Step-by-step solution

Explanation of the ideal gas law

The pressure, volume, temperature, and the number of moles of a gas are all related by the ideal gas law.

It is given as

\({\rm{PV = nRT}}\).

\(\begin{aligned}{}{\rm{P = pressure of the gas}}{\rm{. }}\\{\rm{V = volume of the gas}}{\rm{.}}\\{\rm{ n = number of moles of the gas}}{\rm{.}}\\{\rm{ T = temperature of the gas}}{\rm{. }}\end{aligned}\)

\(\begin{aligned}{}{\rm{R = Ideal gas constant }}\\{\rm{ = 0}}{\rm{.08206\;L atm mo}}{{\rm{l}}^{{\rm{ - 1}}}}{{\rm{K}}^{{\rm{ - 1}}}}\end{aligned}\).

The amount of the gas \({\rm{(n)}}\)is in moles, temperature \({\rm{(T)}}\) is in Kelvin \(\left( {\rm{K}} \right){\rm{,}}\) and pressure \({\rm{(P)}}\)is in atm.

Finding the number of moles of gas

\(\begin{aligned}{}{\rm{Pressure (P) = 0}}{\rm{.462\;atm}}{\rm{.}}\\{\rm{Volume (V) = 73}}{\rm{.3\;mL}}{\rm{. }}\\{\rm{Temperature (T) = ?}}\\{\rm{K amount of gas (n) = 0}}{\rm{.292 moles}}{\rm{. }}\\{\rm{Ideal gas constant (R) = 0}}{\rm{.08206\;L atm mo}}{{\rm{l}}^{{\rm{ - 1}}}}{{\rm{K}}^{{\rm{ - 1}}}}{\rm{n}}\end{aligned}\)

However, the given volume is in \({\rm{mL}}\)but not in \({\rm{L}}\).

So, we have to convert the volume in\({\rm{mL}}\)to\({\rm{L}}\).

We have\({\rm{1mL = 0}}{\rm{.001L}}\).

So,

\(\begin{aligned}{}{\rm{73}}{\rm{.3\;mL = (73}}{\rm{.3) \times (0}}{\rm{.001)L}}\\{\rm{ = 0}}{\rm{.0733\;L}}\end{aligned}\).

Converting \({\rm{0}}{\rm{.292}}\)grams of \({{\rm{I}}_{\rm{2}}}\)gas to the no. of moles of \({{\rm{I}}_{\rm{2}}}\) gas:

The atomic weight of iodine \((I) = 126.9\)grams.

So, the molecular weight of \({{\rm{I}}_{\rm{2}}}{\rm{ = (2) \times (126}}{\rm{.9)}}\) grams.

\({\rm{I = 253}}{\rm{.8}}\)grams.

So,\({\rm{253}}{\rm{.8}}\)grams of \({I_{\rm{2}}}{\rm{ = 1}}\)mole of\({{\rm{I}}_{\rm{2}}}\).

So, the no. of moles of gas of \({{\rm{I}}_{\rm{2}}}{\rm{(n) = }}\left( {{\rm{0}}{\rm{.292}}} \right.\) grams of \(\left. {{{\rm{I}}_{\rm{2}}}} \right){\rm{ \times }}\left( {\frac{{{\rm{1 mole }}}}{{{\rm{253}}{\rm{.8 grams of }}{{\rm{I}}_{\rm{2}}}}}} \right)\)

\({\rm{ = 0}}{\rm{.0011}}\)moles of \({{\rm{I}}_{\rm{2}}}\).

Evaluating the temperature

\({\rm{PV = nRT}}\).

\(\begin{aligned}{}{\rm{ Temperature (T) = }}\frac{{{\rm{PV}}}}{{{\rm{nR}}}}\\{\rm{ = }}\frac{{{\rm{(0}}{\rm{.462) \times (0}}{\rm{.0733)}}}}{{{\rm{(0}}{\rm{.0011) \times (0}}{\rm{.08206)}}}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.0338}}}}{{{\rm{0}}{\rm{.00009}}}}\\{\rm{ = 375}}{\rm{.55\;K}}{\rm{.}}\end{aligned}\)

Hence, the temperature is\({\rm{(T) = 375}}{\rm{.55\;K}}{\rm{.}}\)