Chemistry Visualized · States of Matter
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Ideal Gas Equation
The Ideal Gas Law unifies Boyle's, Charles', and Avogadro's laws: PV = nRT. Real gases deviate at high pressure and low temperature.
Variables
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P
2.45
V
10.0
n
1.00
T
298
PV = nRT
2.45×10.0=1.0×0.0821×298
✔ Equation Balanced
Universal Gas Constant (R) Values
0.0821L·atm / (mol·K)8.314J / (mol·K)0.0831L·bar / (mol·K)1.987cal / (mol·K)
Gas Chamber
Mol14
KE1.0×
°C25.0°C
P vs V (Isotherm)
Compressibility (Z vs P)
Teacher's Desk
Boyle's Law
Boyle's Law states that at a constant temperature, the pressure of a fixed mass of an ideal gas is inversely proportional to its volume (P ∝ 1/V or PV = k).
PV = const.
P₁V₁ = P₂V₂ = P₃V₃
PV = K
log P + log V = log K
log P=log K+log (1/V)y=c+mx
Molecular Note: The molecular explanation of the law is based on the fact that the pressure exerted by a gas arises from the impact of its particles on the walls of the vessel. If the volume is halved, the density of particles is doubled, and so twice as many particles strike the walls in a given period of time. The average force exerted by the gas is therefore doubled, and so, is the pressure it exerts.
System State
P
2.45
V
10.0
T
298
P₁V₁ = P₂V₂
2.45×10.0=24.50(constant k)
Isotherm Families (P vs V)
Click a temperature below to force the system state to that isotherm.
Graphical Transformations
These standard curves frequently appear in multiple-choice questions to test your understanding of proportionality constants.
Pressure (P) vs 1 / Volume (1/V) [Interactive]
PV vs Pressure (P) [Interactive]
Pressure (P) vs 1 / Volume (1/V) [Static Reference]
T₃ > T₂ > T₁
Pressure (P) vs Volume (V) [Static Reference]
T₃ > T₂ > T₁
Application: Solved Examples
Mastering how Boyle's Law is tested in competitive exams.
Example 1Gas Mixing
What will be the pressure of the gaseous mixture when 0.5 L of H₂ at 0.8 bar and 2.0 L of dioxygen at 0.7 bar are introduced in a 1 L vessel at 27°C?
Solution
For H₂: p₁V₁ = p₂V₂
0.8 × 0.5 = pH₂ × 1 ⇒ pH₂ = 0.4 bar
For O₂: p₁V₁ = p₂V₂
0.7 × 2.0 = pO₂ × 1 ⇒ pO₂ = 1.4 bar
Therefore, ptotal = pH₂ + pO₂ = 0.4 + 1.4 = 1.8 bar
0.8 × 0.5 = pH₂ × 1 ⇒ pH₂ = 0.4 bar
For O₂: p₁V₁ = p₂V₂
0.7 × 2.0 = pO₂ × 1 ⇒ pO₂ = 1.4 bar
Therefore, ptotal = pH₂ + pO₂ = 0.4 + 1.4 = 1.8 bar
Teacher's Insight: Treat each gas completely independently first. Calculate their new partial pressures using Boyle's Law, then simply add them up (Dalton's Law of Partial Pressures).
Example 2Fractional Change
A gas is present at a pressure of 2 atm. What should be the increase in pressure so that the volume of the gas can be decreased to 1/4th of the initial value if the temperature is maintained constant?
Solution
p₁V₁ = p₂V₂ (constant mass & T)
Given: p₁ = 2 atm, V₂ = V₁/4
2 × V₁ = p₂ × (V₁ / 4)
p₂ = 8 atm
Total increase in pressure = 8 - 2 = 6 atm
Given: p₁ = 2 atm, V₂ = V₁/4
2 × V₁ = p₂ × (V₁ / 4)
p₂ = 8 atm
Total increase in pressure = 8 - 2 = 6 atm
Teacher's Insight: Be very careful with wording! Examiners love asking for the "increase" or "change" rather than the final value. Always read the last line of the question twice.
Example 3Connected Bulbs
Two glass bulbs A and B at same temperature are connected by a small tube having a stop cock. Bulb A has a volume of 100 cm³ and contained the gas while bulb B was empty. On opening the stop cock, the pressure fell down to 20%. What is the volume of bulb B?
Solution
V₁ = 100 cm³, p₁ = 100 atm (let initial be 100%)
p₂ = 20 atm (fell down to 20%)
V₂ = (p₁V₁) / p₂ = (100 × 100) / 20 = 500 cm³
Total Volume (V₂) = VA + VB
Volume of B = 500 - 100 = 400 cm³
p₂ = 20 atm (fell down to 20%)
V₂ = (p₁V₁) / p₂ = (100 × 100) / 20 = 500 cm³
Total Volume (V₂) = VA + VB
Volume of B = 500 - 100 = 400 cm³
Teacher's Insight: The trick here is that the final volume (V₂) is the combined volume of both bulbs. Subtract the original bulb's volume to find the unknown bulb's volume.
Example 4Composite Systems
Consider a composite system at 300K with 3 compartments separated by barriers. Assuming ideal gas behavior, calculate the total pressure if the barriers separating the compartments are removed.
He
2 L
1.50 atm
Ne
4 L
2.50 atm
Xe
1 L
1.00 atm
Solution
p₁V₁ + p₂V₂ + p₃V₃ = pTVtotal
Vtotal = 2 + 4 + 1 = 7 L
(2 × 1.5) + (4 × 2.5) + (1 × 1) = 7 × pT
3 + 10 + 1 = 7 × pT
14 = 7 × pT ⇒ pT = 2 atm
Vtotal = 2 + 4 + 1 = 7 L
(2 × 1.5) + (4 × 2.5) + (1 × 1) = 7 × pT
3 + 10 + 1 = 7 × pT
14 = 7 × pT ⇒ pT = 2 atm
Teacher's Insight: Since PV is proportional to moles (n) at constant T, summing up the PV products of individual compartments gives you the PV product of the final combined system.
Test Your Understanding
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Solution & Insight
Charles' Law Controls
At constant P and n: V ∝ T.
V
12.2
T
298
P
2.00
V₁ / T₁ = V₂ / T₂
12.2/298=0.041(constant k₂)
Key Observations: All lines for different pressures intersect at -273.15 °C (0 K), known as Absolute Zero. At this theoretical temperature, the volume of an ideal gas becomes zero.
Note: Charles's law is most strictly obeyed by all gases at high temperatures and very low pressures.
Note: Charles's law is most strictly obeyed by all gases at high temperatures and very low pressures.
Isobar Families (Volume vs Temp °C)
Lower pressures result in steeper slopes (p₁ < p₂ < p₃ < p₄). Click to change P.
Volume vs Temp (K)
log V vs log T (45° slope)
Application: Solved Examples
Mastering how Charles' Law is tested in competitive exams.
Example 1Fraction Expelled
A student forgot to add the reaction mixture to the round bottomed flask at 27°C but put it on the flame. After a lapse of time, he realized his mistake. Using a pyrometer, he found the temperature of the flask was 477°C. What fraction of the air would have been expelled out?
Solution
Let volume of air at 27°C (300 K) be V₁ and at 477°C (750 K) be V₂.
According to Charles' Law: V₁ / T₁ = V₂ / T₂
Fraction expelled = (V₂ - V₁) / V₂ = 1 - (V₁ / V₂)
Substituting V₁ / V₂ = T₁ / T₂, we get:
Fraction expelled = 1 - (300 / 750) = 450 / 750 = 0.6 (or 3/5)
According to Charles' Law: V₁ / T₁ = V₂ / T₂
Fraction expelled = (V₂ - V₁) / V₂ = 1 - (V₁ / V₂)
Substituting V₁ / V₂ = T₁ / T₂, we get:
Fraction expelled = 1 - (300 / 750) = 450 / 750 = 0.6 (or 3/5)
Teacher's Insight: The volume of the flask itself is constant, but the volume of the gas sample expands. The fraction of gas expelled is based on the new, expanded volume (V₂) trying to fit back into the original rigid container (V₁).
Example 2Connected Vessels
Two vessels are connected by a valve of negligible volume. Container (I) has 2.8 g of N₂ at temperature T₁. Container (II) is completely evacuated. Container (I) is heated to T₂ while (II) is maintained at T₂/3. Volume of (I) is half of (II). If the valve is opened, what is the weight ratio of N₂ in both vessels (w₁/w₂)?
Solution
Since the valve is open, the final pressure P is equalized in both vessels.
From PV=nRT, n ∝ V/T (which implies w ∝ V/T).
w₁ / w₂ = (V₁ / T₁) / (V₂ / T₂)
Given: V₁ = V₂ / 2 and Temp for II = T₂ / 3
w₁ / w₂ = (V₂/2) / T₂ / [V₂ / (T₂/3)]
w₁ / w₂ = 1 / (2 × 3) = 1 / 6
From PV=nRT, n ∝ V/T (which implies w ∝ V/T).
w₁ / w₂ = (V₁ / T₁) / (V₂ / T₂)
Given: V₁ = V₂ / 2 and Temp for II = T₂ / 3
w₁ / w₂ = (V₂/2) / T₂ / [V₂ / (T₂/3)]
w₁ / w₂ = 1 / (2 × 3) = 1 / 6
Teacher's Insight: Don't calculate actual pressures! When containers are connected, gas flows until pressures are equal. Since R is constant, the moles (or weight) distributed in each compartment is simply proportional to V/T for that specific compartment.
Example 3Graphical Derivation
For a closed (not rigid) container containing an ideal gas fitted with a movable piston operating such that pressure remains constant. Draw and explain the graph that represents the variation of log V vs log T.
Solution
From Charles' Law: V = k T
Taking log on both sides: log V = log T + log k
Comparing with the straight line equation y = mx + c:
y = log V, x = log T, m = 1, c = log k
Slope m = tan(θ) = 1 ⇒ θ = 45°
Taking log on both sides: log V = log T + log k
Comparing with the straight line equation y = mx + c:
y = log V, x = log T, m = 1, c = log k
Slope m = tan(θ) = 1 ⇒ θ = 45°
Teacher's Insight: Logarithmic graphs of power laws always yield straight lines. The exponent of the variable becomes the slope. Since V is proportional to T¹, the slope is exactly 1 (an angle of 45 degrees). Look at the graph above to visualize this!
Test Your Understanding
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Solution & Insight
Graham's Law of Diffusion & Effusion
The rate at which a gas diffuses or effuses is dependent on its macroscopic properties. According to kinetic theory, the rate is directly proportional to the pressure and the area of the opening, and inversely proportional to the square root of the temperature and its molar mass.
Diffusion is the mixing of gas molecules by random motion under conditions where molecular collisions occur (e.g., two gases mixing in a room).
Effusion is the escape of a gas through a pinhole into a vacuum without molecular collisions.
Gas 1 Controls
P
5.0
T
300
M
32
A
20
Effusion Chamber
Escaped0
Teacher's Insight: In Effusion mode, particles that escape the hole are teleported back to the left chamber to maintain a constant pressure, allowing you to visually observe a steady macroscopic flow rate. Heavy molecules ($M$) move slower, while higher temperatures ($T$) increase velocity.
Rate∝P × A√(T × M)
P = Pressure
A = Area of pinhole
T = Temperature
M = Molecular Wt.
Rate==dx/dt
dist. travelled by gas
Time taken
==dV/dt
vol. of gas escaped
Time taken
==dn/dt
mole of gas escaped
Time taken
Relative Rate Factor = 1.00x
Gas States Concept Matrix
A structured breakdown of empirical laws, microscopic theories, and real gas deviations.
Macroscopic Laws
Boyle's Law
P ∝ 1/V ⇒ P₁V₁ = P₂V₂
At a constant temperature (T) and moles (n), the pressure of a gas is inversely proportional to its volume.
Charles' & Gay-Lussac's
V₁/T₁ = V₂/T₂ | P₁/T₁ = P₂/T₂
Charles': Volume is directly proportional to absolute Temp (constant P).
Gay-Lussac's: Pressure is directly proportional to absolute Temp (constant V).
Gay-Lussac's: Pressure is directly proportional to absolute Temp (constant V).
Dalton's Law of Partial Pressures
Ptotal = p₁ + p₂ + p₃...
Total pressure is the sum of partial pressures. Partial pressure is calculated as: p₁ = x₁ · Ptotal (where x is mole fraction).
Microscopic Theory
Kinetic Molecular Theory (KMT)
- Actual volume of molecules is negligible (point masses).
- No intermolecular forces of attraction or repulsion.
- Continuous, random motion with perfectly elastic collisions.
- Gas pressure is caused by collisions with container walls.
Kinetic Gas Equation
PV = ⅓ mnu²
Relates macroscopic properties (PV) to microscopic motion (u = root mean square speed).
Average K.E. / mole = ¾ RT
Average K.E. / mole = ¾ RT
Molecular Velocities
Urms = √(3RT/M)
Urms (Root Mean Square) > Uav (Average) > Ump (Most Probable)
Ratio of Ump : Uav : Urms = 1 : 1.128 : 1.224
Ratio of Ump : Uav : Urms = 1 : 1.128 : 1.224
Real Deviations
Van der Waals Equation
(P + an²/V²)(V - nb) = nRT
Corrects ideal gas behavior for real conditions:
'a' = measure of intermolecular attractive forces.
'b' = effective volume occupied by gas molecules.
'a' = measure of intermolecular attractive forces.
'b' = effective volume occupied by gas molecules.
Compressibility Factor (Z)
Z = PV / nRT
Z = 1: Ideal behavior.
Z < 1: Negative deviation (attractive forces dominate, easily compressible).
Z > 1: Positive deviation (repulsive forces dominate, hard to compress).
Z < 1: Negative deviation (attractive forces dominate, easily compressible).
Z > 1: Positive deviation (repulsive forces dominate, hard to compress).
Critical Constants
Tc = 8a / 27Rb
Tc: Highest temp at which gas can be liquefied.
Pc = a / 27b² (Pressure at Tc)
Vc = 3b (Volume of 1 mol at Tc)
Pc = a / 27b² (Pressure at Tc)
Vc = 3b (Volume of 1 mol at Tc)