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ChemistryEdu Logo Gaseous State | Ideal Gas Equation#

Ideal Gas Equation#

  • By Boyle's Law, we know that for a given mass of a gas, pressure is inversely proportional to volume at constant temperature.
\[P ∝ {1 \over V}\]
  • By Charle's Law, we know that for a given mass of a gas, volume occupied by the gas is directly proportional to temperature at constant pressure.
\[V ∝ T\]
  • By Gay Lussac's Law, we know that for a given mass of a gas, pressure is directly proportional to temperature at constant volume.
\[P ∝ T\]
  • By Avogadro's Law, we know that volume occupied by a gas is directly proportional to its moles at constant temperature and pressure.
\[V ∝ n\]
  • Combining all above four equations, we obtain the ideal gas equation:
\[PV ∝ nT\]
\[PV = nRT\]

Here, R = Universal Gas Constant

Values of R in different units

\[R = 0.0821\ atm\ L\ mol^{-1}\]
\[R = 8.314\ J\ mol^{-1}\ K^{-1}\]
\[R = 1.98 ≈ 2\ calorie\ mol^{-1}\ K^{-1}\]
  • Note that \(1\ calorie = 4.184\ J\)

Questions#

An ideal gas at 300 K is kept in an open container of volume 5 L. When temperature is raised to 500 K keeping pressure constant, calculate the percentage of moles of gas molecules escaped.

\[PV = nRT\]

Here, PV = constant because P and V both are constants.

\[Constant = nRT\]
\[n ∝ {1 \over T}\]

Let initial and final moles be ni and nf respectively and let initial and final temperatures be Ti and Tf.

\[ {n_i \over n_f} = {T_f \over T_i} \]
\[ {n_i \over n_f} = {500 \over 300} \]
\[ {n_i \over n_f} = {5 \over 3} \]
\[ {n_f \over n_i} = {3 \over 5} \]

Percentage of moles of gas escaped:

\[Percentage\ of\ moles\ of\ gas\ escaped = {n_i - n_f \over n_i} \times 100\]
\[Percentage\ of\ moles\ of\ gas\ escaped = {1 - {n_f \over n_i}} \times 100\]
\[Percentage\ of\ moles\ of\ gas\ escaped = {1 - {3 \over 5}} \times 100 = 40\]

A water bubble on the surface has volume = 4V. When it reaches at a depth of 'h', its volume becomes V. If 10 m water = 1 atm, calculate the value of 'h'.

Let initial pressure be \(P_1\), final pressure be \(P_2\), initial volume be \(V_1\) and final volume be \(V_2\).

\[ {P_1 \over P_2} = {V_2 \over V_1}\]
\[ {1\ atm + h \over 1\ atm} = {4V \over V}\]
\[ {10 + h \over 10} = 4\]
\[ h = 30\ m\]

Find the common pressure when the stop cork is removed. The scenario is shown below:

Let the common pressure attained when the stop cork is removed be P.

\[Initial\ moles = Final\ moles\]
\[ {P_A V \over RT} + {P_B V \over RT} = {P(V + V) \over RT} \]
\[ P_A + P_B = 2P \]
\[ 3 + 6 = 2P \]
\[ P = 4.5\ atm \]