How many grams of gas are present in each of the following cases?

(a)0.1L of CO₂at 307 torr and \({\rm{2}}{{\rm{6}}^{\rm{^\circ }}}{\rm{C}}\)

(b) 8.75L of C₂H₄, at 378.3KPa and 483K

(c) \({\rm{221\;mL}}\)of Ar at 0.23 torr and -54⁰C

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.

a)

Given,

\(\begin{aligned}{}{\rm{Volume (V) = 0}}{\rm{.1\;L}}{\rm{.}}\\{\rm{Pressure (P) = 307 torr}}{\rm{. }}\\{\rm{Temperature (T) = 2}}{{\rm{6}}^{\rm{^\circ }}}{\rm{. }}\\{\rm{Amount of gas (n) = ? moles}}{\rm{. }}\end{aligned}\)

However, temperatureismeasuredin celsiusrather than kelvins (absolute scale), and pressure is measured in torrs rather than atm.

As a result, we must convert the temperature from degrees celsiusto kelvins.

We have\({{\rm{0}}^{\rm{^\circ }}}{\rm{ = (0) + 273}}{\rm{.15K}}\).

So,

\(\begin{aligned}{}{\rm{2}}{{\rm{6}}^{\rm{^\circ }}}{\rm{ = (26) + 273}}{\rm{.15K}}\\{\rm{ = 299}}{\rm{.15K}}\end{aligned}\).

Converting pressure in torr into atm:

We have\({\rm{1 torr = 0}}{\rm{.0013}}\)atm.

So,

\(\begin{aligned}{}{\rm{307 torr = (307) \times (0}}{\rm{.0013) atm}}{\rm{. }}\\{\rm{ = 0}}{\rm{.403\;atm}}{\rm{.}}\end{aligned}\).

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

The no. of moles of gas \({\rm{(n) = }}\frac{{{\rm{PV}}}}{{{\rm{RT}}}}\)

\(\begin{aligned}{}{\rm{ = }}\frac{{{\rm{(0}}{\rm{.403) \times (0}}{\rm{.1)}}}}{{{\rm{(0}}{\rm{.08206) \times (299}}{\rm{.15)}}}}{\rm{mol}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.0403}}}}{{{\rm{24}}{\rm{.55}}}}\\{\rm{ = 0}}{\rm{.00164 moles}}{\rm{. }}\end{aligned}\)

Converting \({\rm{0}}{\rm{.00164}}\)moles of \({\rm{C}}{{\rm{O}}_2}\) into grams of\({\rm{C}}{{\rm{O}}_{\rm{2}}}\):

The atomic weight of carbon \(\left( {\rm{C}} \right){\rm{ = 12}}\) grams.

The atomic weight of oxygen \(\left( {\rm{O}} \right){\rm{ = 16}}\) grams.

So, the molecular weight of \({\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{ = (12) + (2) \times (16)}}\) grams.

\(\begin{aligned}{}{\rm{ = 12 + 32}}\\{\rm{ = 44 grams}}\end{aligned}\).

So,\({\rm{1}}\)mole of \({\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{ = 44}}\) grams of\({\rm{C}}{{\rm{O}}_{\rm{2}}}\).

\({\rm{0}}{\rm{.0016}}\)moles of \({\rm{C}}{{\rm{O}}_{\rm{2}}}{\rm{ = (0}}{\rm{.00164) \times (44)}}\) grams of\({\rm{C}}{{\rm{O}}_{\rm{2}}}\).

\({\rm{ = 0}}{\rm{.072}}\)grams of\({\rm{C}}{{\rm{O}}_2}\).

\( = 7.2 \times {10^{ - 2}}\)grams of\({\rm{C}}{{\rm{O}}_2}\).

After evaluating, we get\(7.2 \times {10^{ - 2}}\)grams of \({\rm{C}}{{\rm{O}}_2}\).

Given:

\(\begin{aligned}{}{\rm{ Volume (V) = 8}}{\rm{.75\;L}}{\rm{. }}\\{\rm{Pressure (P) = 378}}{\rm{.3kPa}}{\rm{. }}\\{\rm{Temperature (T) = 483\;K}}{\rm{.}}\\{\rm{ Amount of gas (n) = ? moles}}{\rm{. }}\end{aligned}\)

Converting pressure in \({\rm{kPa}}\)into atm:

We have\({\rm{1kPa = 0}}{\rm{.0098\;atm}}\).

So,

\(\begin{aligned}{}{\rm{378}}{\rm{.3kPa = (378}}{\rm{.3) \times (0}}{\rm{.0098)atm}}{\rm{.}}\\{\rm{ = 3}}{\rm{.73\;atm}}\end{aligned}\).

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

The no. of moles of gas \({\rm{(n) = }}\frac{{{\rm{PV}}}}{{{\rm{RT}}}}\)

\(\begin{aligned}{}{\rm{ = }}\frac{{{\rm{(3}}{\rm{.73) \times (8}}{\rm{.75)}}}}{{{\rm{(0}}{\rm{.08206) \times (483)}}}}{\rm{ moles}}{\rm{. }}\\{\rm{ = }}\frac{{{\rm{32}}{\rm{.6375}}}}{{{\rm{39}}{\rm{.6349}}}}\\{\rm{ = 0}}{\rm{.82 moles}}{\rm{. }}\end{aligned}\)

Converting \({\rm{0}}{\rm{.82molesof}}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}\)into grams of \({{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}\):

The atomic weight of carbon (C) \({\rm{ = 12}}\) grams.

The atomic weight of hydrogen \({\rm{(H) = 1}}\)gram.

So, the molecular weight of \({{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{ = (2) \times (12) + (4) \times (1)}}\)grams.

\({\rm{ = 24 + 4}}\)

\({\rm{ = 28}}\)grams.

So, 1 mole of \({{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{ = 28}}\) grams of \({{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}\).

\({\rm{0}}{\rm{.82}}\)moles of \({{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{ = (0}}{\rm{.82) \times (28)}}\) grams of \({{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}\).

\({\rm{ = 22}}{\rm{.96}}\)grams of \({{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}\).

After evaluating, we get\({\rm{22}}{\rm{.96}}\)grams of \({{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}\).

Given:

\(\begin{aligned}{}{\rm{ Volume (V) = 221\;mL}}{\rm{. }}\\{\rm{Pressure (P) = 0}}{\rm{.23 torr}}{\rm{. }}\\{\rm{Temperature (T) = - 5}}{{\rm{4}}^{\rm{^\circ }}}{\rm{. }}\\{\rm{Amount of gas (n) = ? moles}}{\rm{. }}\end{aligned}\)

However, temperatureisprovided in degrees celsiusrather than kelvins (absolute scale), pressure is given in torrs rather than atm, and volume is given in mL rather than L.

As a result, we must convert the temperature from degrees celsiusto kelvins. We have\({{\rm{0}}^{\rm{^\circ }}}{\rm{ = (0) + 273}}{\rm{.15K}}\)

So,

\(\begin{aligned}{}{\rm{ - 5}}{{\rm{4}}^{\rm{^\circ }}}{\rm{ = ( - 54) + 273}}{\rm{.15K}}\\{\rm{ = 219}}{\rm{.15K}}\end{aligned}\).

Converting pressure in torr into atm:

We have 1 torr \({\rm{ = 0}}{\rm{.0013\;atm}}{\rm{.}}\)

So,

\(\begin{aligned}{}{\rm{0}}{\rm{.23 torr = (0}}{\rm{.23) \times (0}}{\rm{.0013) atm }}\\{\rm{ = 0}}{\rm{.00029\;atm}}\end{aligned}\).

Converting volume in \({\rm{mL}}\)into \({\rm{L}}{\rm{.}}\):

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

So,

\(\begin{aligned}{}{\rm{221mL = (221) \times (0}}{\rm{.001)L}}{\rm{.}}\\{\rm{ = 0}}{\rm{.221L}}\\{\rm{.PV = nRT}}\end{aligned}\).

\(\begin{aligned}{}{\rm{The no}}{\rm{. of moles of gas(n) = }}\frac{{{\rm{PV}}}}{{{\rm{RT}}}}\\{\rm{ = }}\frac{{{\rm{(0}}{\rm{.00029) \times (0}}{\rm{.221)}}}}{{{\rm{(0}}{\rm{.08206) \times (219}}{\rm{.15)}}}}{\rm{ moles}}{\rm{. }}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.000064}}}}{{{\rm{17}}{\rm{.983}}}}\\{\rm{ = 0}}{\rm{.0000035 moles}}{\rm{. }}\end{aligned}\)

Converting 0.0000035 moles of Ar into grams of Ar:

The molecular weight of argon \({\rm{(Ar) = 40}}\)grams.

\(\begin{aligned}{}{\rm{ i}}{\rm{.e}}{\rm{. 1 mole of (Ar) = 40 grams of }}{{\rm{C}}_{\rm{2}}}{{\rm{H}}_{\rm{4}}}{\rm{.}}\\{\rm{0}}{\rm{.0000035 moles of (Ar) = (0}}{\rm{.0000035) \times (40) grams of Ar}}{\rm{. }}\\{\rm{ = 0}}{\rm{.000014 grams of Ar}}{\rm{. }}\\{\rm{ = 1}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}{\rm{ grams of Ar}}{\rm{. }}\end{aligned}\)

After evaluating, we get the required number as\({\rm{1}}{\rm{.4 \times 1}}{{\rm{0}}^{{\rm{ - 4}}}}\)grams of Ar.