Chemistry, 25.07.2019 11:20, graceaowen

The pressure in a 12.2 l vessel that contains 2.34 g of carbon dioxide, 1.73 g of sulfur dioxide, and 3.33 g of argon, all at 42°c is torr.

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The pressure in a 12.2 L vessel that contains 2.34 g of ${\text{C}}{{\text{O}}_2}$ , 1.73 g of ${\text{S}}{{\text{O}}_{\text{2}}}$ and 3.33 g of Ar is $\boxed{{\text{262}}{\text{.6 torr}}}$ .

Further explanation:

An ideal gas is a hypothetical gas that is composed of a large number of randomly moving particles that are supposed to have perfectly elastic collisions among themselves. It is just a theoretical concept and practically no such gas exists. But gases tend to behave almost ideally at a higher temperature and lower pressure.

Ideal gas law is the equation of state for any hypothetical gas. The expression for the ideal gas equation is as follows:

${\text{PV}}={\text{nRT}}$ …… (1)

Here,

P is the pressure of the gas.

V is the volume of gas.

T is the absolute temperature of the gas.

n is the number of moles of gas.

R is the universal gas constant.

The formula to calculate the moles of substance is as follows:

${\text{Moles of substance}} = \frac{{{\text{Given mass of substance}}}}{{{\text{Molar mass of substance}}}}$ ...... (2)

Substitute 2.34 g for the given mass and 44 g/mol for the molar mass in equation (2) to calculate the moles of ${\text{C}}{{\text{O}}_2}$

$\begin{aligned}{\text{Moles of C}}{{\text{O}}_2}&=\left({{\text{2}}{\text{.34 g}}} \right)\left( {\frac{{{\text{1 mol}}}}{{{\text{44 g}}}}} \right)\\&=0.053{\text{1 mol}}\\&\approx {\text{0}}{\text{.053 mol}} \\ \end{aligned}$

Substitute 1.73 g for the given mass and 64 g/mol for the molar mass in equation (2) to calculate the moles of ${\text{S}}{{\text{O}}_2}$

$\begin{aligned}{\text{Moles of S}}{{\text{O}}_2}&=\left({{\text{1}}{\text{.73 g}}} \right)\left( {\frac{{{\text{1 mol}}}}{{{\text{64 g}}}}} \right)\\&=0.027{\text{ mol}} \\ \end{aligned}$

Substitute 3.33 g for the given mass and 40 g/mol for the molar mass in equation (2) to calculate the moles of Ar.

$\begin{aligned}{\text{Moles of S}}{{\text{O}}_2}&=\left({{\text{3}}{\text{.33 g}}} \right)\left( {\frac{{{\text{1 mol}}}}{{{\text{40 g}}}}} \right)\\&=0.0832{\text{ mol}}\\&\approx {\text{0}}{\text{.083 mol}} \\ \end{aligned}$

The formula to calculate the total moles of gas in the vessel is as follows:

${\text{Moles of gas}} = {\text{Moles of C}}{{\text{O}}_{\text{2}}} + {\text{Moles of S}}{{\text{O}}_{\text{2}}} + {\text{Moles of Ar}}$ ...... (3)

Substitute 0.053 mol for the moles of ${\text{C}}{{\text{O}}_{\text{2}}}$ , 0.027 mol for the moles of ${\text{S}}{{\text{O}}_{\text{2}}}$ and 0.083 mol for the moles of Ar in equation (3).

$\begin{aligned}{\text{Moles of gas }}\left( {\text{n}} \right)&={\text{0}}{\text{.053 mol}} + {\text{0}}{\text{.027 mol}} + {\text{0}}{\text{.083 mol}}\\&={\text{0}}{\text{.163 mol}} \\ \end{aligned}$

Rearrange equation (1) to calculate the pressure of the gas.

${\text{P}} = \frac{{{\text{nRT}}}}{{\text{V}}}$ ...... (4)

Substitute 0.163 for n, 12.2 L for V, ${\text{42}}\;^\circ {\text{C}}$ for T and $62.364\;{\text{L}} \cdot {\text{torr}} \cdot {\text{mo}}{{\text{l}}^{ - 1}} \cdot {{\text{K}}^{ - 1}}$ for R in equation (4).

$\begin{aligned}{\text{P}}&=\frac{{\left( {{\text{0}}{\text{.163 mol}}} \right)\left( {62.364\;{\text{L}} \cdot {\text{torr}} \cdot {\text{mo}}{{\text{l}}^{ - 1}} \cdot {{\text{K}}^{ - 1}}} \right)\left( {42 + 273.15} \right)\;{\text{K}}}}{{{\text{12}}{\text{.2 L}}}}\\&=262.590{\text{5 torr}} \\&\approx 262.6\;{\text{torr}} \\ \end{aligned}$

Therefore the pressure in the vessel is 262.6 torr.

Learn more:

1. Which statement is true for Boyle’s law:

2. Calculation of volume of gas:

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ideal gas equation

Keywords: pressure, ideal gas equation, P, V, n, R, T, 0.163 mol, CO2, SO2, Ar, moles, molar mass, temperature, pressure, number of moles, moles of CO2, moles of SO2, moles of Ar, 0.053 mol, 0.027 mol, 0.083 mol, 262.6 torr.

answered: Guest

quartz

explanation: