Question

How many grams of \[CO\left( g \right)\] are there in \[745{\text{ }}ml\] of the gas at \[1.03{\text{ }}atm\] and \[{36^o}\] Celsius?

Answer 1

Hint: The ideal gas law, likewise called the general gas equation, is the condition of state of a theoretical ideal gas. It is a decent estimation of the conduct of numerous gases under numerous conditions, in spite of the fact that it has a few constraints. The ideal gas law is frequently written in an observational formula:
\[PV = nRT\]
where \[P\], \[V\] and \[T\] are the pressure, volume and temperature; \[n\] is the measure of substance; and \[R\] is the ideal gas constant.

Complete step by step answer:
The condition of a measure of gas is controlled by its pressure, volume, and temperature. The advanced type of the condition relates these just in two principal structures. The temperature utilized in the condition of state is a flat-out temperature: the proper SI unit is the kelvin.
This is a clear utilization of the ideal gas law equation:
\[PV = nRT\]
Where:
\[P\] - the pressure of the gas, for your situation equivalent to \[1.03{\text{ }}atm\]
\[V\] - the volume it possesses, for your situation \[745{\text{ }}mL\]
\[n\] - the quantity of moles of gas
\[R\] - the universal gas constant, typically given as $0.082\dfrac{{{\text{atm}} \cdot {\text{L}}}}{{{\text{mol}} \cdot {\text{K}}}}$
\[T\] - the temperature of the gas, communicated in Kelvin
In SI units, \[p\] is estimated in pascals, \[V\] is estimated in cubic meters, \[n\] is estimated in moles, and \[T\] in kelvins. \[R\] has the worth \[8.314{\text{ }}J/\left( {K\cdot mol} \right){\text{ }} \approx {\text{ }}2{\text{ }}cal/\left( {K\cdot mol} \right)\], or \[0.0821{\text{ }}l.atm/\left( {mol\cdot} \right)\].
Notice that the inquiry furnishes you with all you require to locate the quantity of moles of gas. Since you know the character of the gas, you would then be able to utilize its molar mass to locate the mass of the example. The principal activity here is to ensure that the units coordinate those utilized in the universal gas constant. A speedy correlation shows that you need to change the volume from milliliters over to liters and the temperature from degrees Celsius to Kelvin.
In this way, plug in these quantities and tackle the ideal gas law condition for \[n\]
$n = \dfrac{{PV}}{{RT}}$
$n = \dfrac{{1.03\not{{{\text{atm}}}} \cdot 745 \cdot {{10}^{ - 3}}\not{{\text{L}}}}}{{0.082\dfrac{{\not{{{\text{atm}}}} \cdot \not{{\text{L}}}}}{{{\text{mol}} \cdot \not{{\text{K}}}}} \cdot \left( {273.15 + 36} \right)\not{{\text{K}}}}} = {\text{0}}{\text{.03027 moles}}$
Molar mass of Carbon monoxide, \[CO\] is \[28.01{\text{ }}g/mol\], it follows that your example will have a mass of
$0.3027\not{{{\text{moles}}}} \cdot \dfrac

{{{\text{28}}{\text{.01 g}}}}{{1\not{{{\text{mole}}}}}} = {\text{0}}{\text{.84786 g}}$
Adjusted to two sig figs, the quantity of sig figs you have for the temperature of the gas, the appropriate response will be
$\left[ {m = {\text{0}}{\text{.85 g}}} \right]$

Note: The ideal gas law relates the four free actual properties of a gas whenever. The ideal gas law can be utilized in stoichiometry issues in which substance responses include gases. Standard temperature and pressure (STP) are a valuable arrangement of benchmark conditions to look at different properties of gases. At STP, gases have a volume of \[22.4{\text{ }}L\] per mole. The ideal gas law can be utilized to decide densities of gases.