Gaseous state

Question

Give answers to the following

reagansimiyu · Answer

1. derive the si unit of pressure
pressure (p) is defined as force per unit area:
p = f / a
the si unit of force (f) is the newton (n), and area (a) is in square meters (m²).
from newton's second law: f = ma
- mass (m) = kg
- acceleration (a) = m/s²
→ 1 n = 1 kg·m/s²
thus,
p = 1 n/m² = (1 kg·m/s²) / m² = 1 kg·m⁻¹·s⁻²
hence, the si unit of pressure is the pascal (pa): 1 pa = 1 kg·m⁻¹·s⁻²
2. define the term 'pressure' of a gas in terms of two different equations
definition 1: p = f / a
- pressure is the force applied per unit area.
definition 2 (kinetic theory): p = (1/3)ρc²
- ρ = gas density
- c = root-mean-square speed of gas molecules
both definitions yield pressure in the same unit: pascal (pa), which is 1 n/m².
3. derive the state equation of ideal gases from empirical gas laws
we combine three laws:
- boyle's law: pv = constant (t and n fixed)
- charles's law: v/t = constant (p and n fixed)
- avogadro's law: v ∝ n (p and t fixed)
combining: v ∝ nt/p → pv ∝ nt
adding constant r (gas constant):
→ pv = nrt
ideal gas law.
4. convert the following:
(a) 0.357 atm to torr:
1 atm = 760 torr → 0.357 × 760 = 271.32 torr
(b) 6.6 × 10² torr to atm:
660 / 760 = 0.868 atm
(c) 147.2 kpa to torr:
1 atm = 101.325 kpa = 760 torr
147.2 × (760 / 101.325) = 1103.8 torr
5. calculate r from helium experiment
given:
- mass = 0.126 g, molar mass of he = 4.003 g/mol → n = 0.0315 mol
- t = 102°c = 375.15 k
- v = 10.0 dm³ = 0.010 m³
- pressure = 103.2 cm h₂o
convert cm h₂o to pa:
- h = 103.2 cm = 1.032 m
- density of h₂o = 1000 kg/m³, g = 9.8 m/s²
p = ρgh = 1000 × 9.8 × 1.032 = 10,117.6 pa
use pv = nrt → r = pv / nt
r = (10,117.6 × 0.010) / (0.0315 × 375.15) = 8.59 j/mol·k
6. molar mass of unknown gas
given:
- p = 715 mmhg = 0.9408 atm
- v = 452 cm³ = 0.452 l
- t = 23.0°c = 296.15 k
- mass = 1.007 g
n = pv / rt = (0.9408 × 0.452) / (0.0821 × 296.15) = 0.0175 mol
molar mass = mass / moles = 1.007 / 0.0175 = 57.5 g/mol
7. moles of co₂ from caco₃
given:
- v = 250 ml = 0.250 l
- p = 1.3 atm
- t = 31°c = 304.15 k
use pv = nrt → n = pv / rt
n = (1.3 × 0.250) / (0.0821 × 304.15) = 0.013 mol
therefore, 0.013 moles of co₂ were produced.
8. pressure in tennis ball
given:
- volume = 144 cm³ = 0.144 l
- mass n₂ = 0.33 g, molar mass = 28.02 → n = 0.0118 mol
- t = 24°c = 297.15 k
p = nrt / v = (0.0118 × 0.0821 × 297.15) / 0.144 = 2.01 atm
9. pressure from 55.0 g ccl₄
given:
- v = 1.25 l, t = 409 k, m = 55.0 g, m = 153.82 g/mol
→ n = 55.0 / 153.82 = 0.3575 mol
p = nrt / v = (0.3575 × 0.0821 × 409) / 1.25 = 9.6 atm ≈ 9.73 bar
→ it cannot exert just 2.00 bar. actual pressure = 9.73 bar
10. argon in sealed vessel
given:
- ar mass = 25.0 g, m = 39.95 → n = 0.625 mol
- v = 1.50 l
- t = 30.0°c = 303.15 k
- p = ?
p = nrt / v = (0.625 × 0.0821 × 303.15) / 1.50 = 10.36 atm ≈ 10.5 bar
upon transfer to a 250 cm³ (0.250 l) container, pressure increases further.

alberteinstein123 · Answer

1. pressure is described as force/area where the si of force is n(newtons) and the si of area is m^2. so you'd be left with n/m^2 which is in fact the si of pressure also known as a pascal

@cesiarojas · Answer

ahfarisa · Answer

let's approach each problem step-by-step, starting with detailed explanations and calculations. i'll keep each answer clear and simple, so you can follow and learn as you go. 1. derive the si of pressure. pressure is defined as force per unit area.

pressure=forcearea ext{pressure} = \frac{ ext{force}}{ ext{area}}

the si unit of force is newton (n). the si unit of area is square meter (m²). so, the si unit of pressure is newton per square meter (n/m²), also known as the pascal (pa).

1 pa=1 n/m21\ ext{pa} = 1\ ext{n/m}^2

2. define the term “pressure” of a gas using two different equations. from these equations show that the overall si unit of pressure is the pascal, pa. (a) using force and area:

p=fap = \frac{f}{a}

f = force (newton, n) a = area (m²) si unit:

nm2=pa\frac{n}{m^2} = ext{pa}

(b) using the ideal gas equation:

pv=nrt ⟹ p=nrtvpv = nrt \implies p = \frac{nrt}{v}

n = moles (mol) r = gas constant (8.314 j·mol⁻¹·k⁻¹) t = temperature (k) v = volume (m³) si unit:

p=(mol)(j\cdotpmol−1⋅k−1)(k)m3=jm3=n⋅mm3=nm2=pap = \frac{( ext{mol}) ( ext{j·mol}^{-1}· ext{k}^{-1})( ext{k})}{ ext{m}^3} = \frac{j}{m^3} = \frac{n·m}{m^3} = \frac{n}{m^2} = ext{pa}

3. from empirical gas laws, derive the state equation of ideal gases. boyle’s law:

pv=kpv = k

(at constant t, n) charles’s law:

vt=k\frac{v}{t} = k

(at constant p, n) avogadro’s law:

vn=k\frac{v}{n} = k

(at constant p, t) combining:

pv=nrtpv = nrt

where r is the universal gas constant. 4. convert the following: (a) 0.357 atm to torr

1 atm=760 torr1~ ext{atm} = 760~ ext{torr}

0.357 atm\times760 torr/atm=271.32 torr0.357~ ext{atm} imes 760~ ext{torr/atm} = 271.32~ ext{torr}

(b) 6.6 × 10⁻² torr to atmospheres

1 atm=760 torr1~ ext{atm} = 760~ ext{torr}

6.6×10−2 torr760 torr/atm=8.68×10−5 atm\frac{6.6 imes 10^{-2}~ ext{torr}}{760~ ext{torr/atm}} = 8.68 imes 10^{-5}~ ext{atm}

(c) 147.2 kpa to torr

1 kpa=7.50062 torr1~ ext{kpa} = 7.50062~ ext{torr}

147.2 kpa\times7.50062 torr/kpa=1104.41 torr147.2~ ext{kpa} imes 7.50062~ ext{torr/kpa} = 1104.41~ ext{torr}

5. calculate the value of r from experimental data. given: volume,

v=10.0 dm3=0.010 m3v = 10.0~ ext{dm}^3 = 0.010~ ext{m}^3

mass he = 0.126 g, molar mass he = 4.00 g/mol ⇒

n=0.126/4=0.0315 moln = 0.126/4 = 0.0315~ ext{mol}

temp = 102°c = 375 k manometer: 103.20 cm h₂o

1 cm h2o=98.0665 pa1~ ext{cm h}_2o = 98.0665~ ext{pa}

p=103.20\times98.0665=10118.45 pap = 103.20 imes 98.0665 = 10118.45~ ext{pa}

plug into

pv=nrtpv = nrt

:

r=pvntr = \frac{pv}{nt}

r=10118.45 pa×0.010 m30.0315 mol×375 kr = \frac{10118.45~ ext{pa} imes 0.010~ ext{m}^3}{0.0315~ ext{mol} imes 375~ ext{k}}

r=101.184511.8125≈8.57 j/mol\cdotpkr = \frac{101.1845}{11.8125} \approx 8.57~ ext{j/mol·k}

6. molar mass of unknown gas. given: mass = 1.007 g pressure = 715 mmhg =

715/760=0.941 atm715/760 = 0.941~ ext{atm}

volume = 452 cm³ = 0.452 l temp = 23°c = 296 k use

pv=nrtpv = nrt

:

n=pvrtn = \frac{pv}{rt}

n=0.941×0.4520.0821×296n = \frac{0.941 imes 0.452}{0.0821 imes 296}

n≈0.42524.31≈0.0175 moln \approx \frac{0.425}{24.31} \approx 0.0175~ ext{mol}

molar mass =

1.0070.0175≈57.5 g/mol\frac{1.007}{0.0175} \approx 57.5~ ext{g/mol}

7. caco₃ decomposition. how many moles of co₂ produced? volume co₂ = 250 ml = 0.250 l p = 1.3 atm t = 31°c = 304 k

n=pvrt=1.3×0.250.0821×304n = \frac{pv}{rt} = \frac{1.3 imes 0.25}{0.0821 imes 304}

n=0.32524.968≈0.013 moln = \frac{0.325}{24.968} \approx 0.013~ ext{mol}

8. tennis ball, n₂ pressure. v = 144 cm³ = 0.144 l mass n₂ = 0.33 g, m = 28 g/mol → n = 0.0118 mol t = 24°c = 297 k

p=nrtvp = \frac{nrt}{v}

p=0.0118×0.0821×2970.144p = \frac{0.0118 imes 0.0821 imes 297}{0.144}

p≈0.28680.144≈1.99 atmp \approx \frac{0.2868}{0.144} \approx 1.99~ ext{atm}

9. density of ccl₄ vapor at 67 torr and 4°c. p = 67 torr = 0.0882 atm t = 4°c = 277 k m = 153.8 g/mol

d=mprtd = \frac{mp}{rt}

d=153.8×0.08820.0821×277d = \frac{153.8 imes 0.0882}{0.0821 imes 277}

d=13.56822.747≈0.597 g/ld = \frac{13.568}{22.747} \approx 0.597~ ext{g/l}

10. argon in vessel: does 25 g in 1.5 dm³ exert 2.0 bar at 30°c as a perfect gas? n = 25/40 = 0.625 mol v = 1.5 l t = 303 k r = 0.08314 l·bar/mol·k

p=nrtvp = \frac{nrt}{v}

p=0.625×0.08314×3031.5p = \frac{0.625 imes 0.08314 imes 303}{1.5}

p=15.7351.5≈10.49 barp = \frac{15.735}{1.5} \approx 10.49~ ext{bar}

this is much greater than 2.0 bar, so it would not behave as a perfect gas at this high pressure (deviations due to real gas behavior). 11. reaction vessel with nh₃ and o₂. given: 120 cm³ nh₃ at 25°c and 750 mmhg 165 cm³ o₂ at 50°c and 635 mmhg both transferred to 0.30 dm³ reaction vessel find: limiting reagent, total pressure after reaction. let’s fully answer question 11 with all the steps. 120 cm³ of nh₃ at 25°c and 750 mmhg is mixed with 165 cm³ of o₂ at 50°c and 635 mmhg, and transferred to a 0.30 dm³ reaction vessel. they react as:

4nh3(g)+5o2(g)→4no(g)+6h2o(g)4 ext{nh}_3 (g) + 5 ext{o}_2 (g) ightarrow 4 ext{no} (g) + 6 ext{h}_2 ext{o} (g)

find: what is the total pressure in the vessel at 150°c after the reaction is over? step 1: calculate the moles of each gas before the reaction a. nh₃

v=120 cm3=0.120 lv = 120~ ext{cm}^3 = 0.120~ ext{l}

p=750 mmhgp = 750~ ext{mmhg}

(

1 atm=760 mmhg1~ ext{atm} = 760~ ext{mmhg}

), so

p=0.987 atmp = 0.987~ ext{atm}

t=25∘c=298 kt = 25^\circ ext{c} = 298~ ext{k}

r=0.0821 l\cdotpatm\cdotpmol−1k−1r = 0.0821~ ext{l·atm·mol}^{-1} ext{k}^{-1}

nnh3=pvrt=0.987×0.1200.0821×298n_{ ext{nh}_3} = \frac{pv}{rt} = \frac{0.987 imes 0.120}{0.0821 imes 298}

nnh3=0.1184424.4758≈0.00484 moln_{ ext{nh}_3} = \frac{0.11844}{24.4758} \approx 0.00484~ ext{mol}

b. o₂

v=165 cm3=0.165 lv = 165~ ext{cm}^3 = 0.165~ ext{l}

p=635 mmhgp = 635~ ext{mmhg}

(

p=0.8368 atmp = 0.8368~ ext{atm}

)

t=50∘c=323 kt = 50^\circ ext{c} = 323~ ext{k}

no2=0.8368×0.1650.0821×323n_{ ext{o}_2} = \frac{0.8368 imes 0.165}{0.0821 imes 323}

no2=0.13807226.5583≈0.00520 moln_{ ext{o}_2} = \frac{0.138072}{26.5583} \approx 0.00520~ ext{mol}

step 2: determine the limiting reagent

4 nh3:5 o24~ ext{nh}_3 : 5~ ext{o}_2

calculate “how much o₂ needed to react with all nh₃”:

no2 needed=0.00484 mol nh3×54=0.00605 mol o2n_{ ext{o}_2~ ext{needed}} = 0.00484~ ext{mol nh}_3 imes \frac{5}{4} = 0.00605~ ext{mol o}_2

but we only have 0.00520 mol o₂ — o₂ is the limiting reagent. step 3: how much nh₃ reacts?

nnh3 used=no2 used×45=0.00520×0.8=0.00416 moln_{ ext{nh}_3~ ext{used}} = n_{ ext{o}_2~ ext{used}} imes \frac{4}{5} = 0.00520 imes 0.8 = 0.00416~ ext{mol}

nh₃ left over:

0.00484-0.00416=0.00068 mol0.00484 - 0.00416 = 0.00068~ ext{mol}

o₂ left over:

0 mol0~ ext{mol}

(all used up) step 4: find the moles of products formed from the reaction: every 5 mol o₂ gives 4 mol no and 6 mol h₂o so,

nno=0.00520 mol o2×45=0.00416 moln_{ ext{no}} = 0.00520~ ext{mol o}_2 imes \frac{4}{5} = 0.00416~ ext{mol}

nh2o=0.00520 mol o2×65=0.00624 moln_{ ext{h}_2o} = 0.00520~ ext{mol o}_2 imes \frac{6}{5} = 0.00624~ ext{mol}

step 5: total moles of gas after reaction nh₃ (left): 0.00068 mol o₂ (left): 0 no (formed): 0.00416 mol h₂o (formed, assume vapor): 0.00624 mol

ntotal=0.00068+0.00416+0.00624=0.01108 moln_{ ext{total}} = 0.00068 + 0.00416 + 0.00624 = 0.01108~ ext{mol}

step 6: find the total pressure at 150°c (423 k) in 0.30 dm³

v=0.30 dm3=0.300 lv = 0.30~ ext{dm}^3 = 0.300~ ext{l}

n=0.01108 moln = 0.01108~ ext{mol}

t=150∘c=423 kt = 150^\circ ext{c} = 423~ ext{k}

r=0.0821 l\cdotpatm\cdotpmol−1k−1r = 0.0821~ ext{l·atm·mol}^{-1} ext{k}^{-1}

p=nrtvp = \frac{nrt}{v}

p=0.01108×0.0821×4230.300p = \frac{0.01108 imes 0.0821 imes 423}{0.300}

p=0.38470.300=1.282 atmp = \frac{0.3847}{0.300} = 1.282~ ext{atm}

final answer: the total pressure in the reaction vessel at 150°c after the reaction is over is approximately

1.28 atm\boxed{1.28~ ext{atm}}

. if you want full solutions for any of the other questions above, just ask!

BrainOnHire · Answer

VarunEvolvixa · Answer

now, let me share some insights from my years of teaching these concepts: the key to mastering gas laws is understanding that they're all interconnected. when students struggle with these problems, it's usually because they're trying to memorize formulas instead of understanding the underlying relationships. for pressure conversions (problem 4), i always tell my students to create a "conversion toolkit" - write down the key relationships like 1 atm = 760 torr = 101.325 kpa and keep them handy. don't try to memorize every possible conversion factor. problem 5 is particularly interesting because it's asking you to experimentally determine the gas constant. this mirrors how early chemists actually discovered these relationships. the slight deviation from the accepted value (8.314 j/mol·k) comes from experimental uncertainty and the fact that real gases don't behave perfectly. for molar mass problems (like #6), remember that the ideal gas law gives you a direct route to moles, and molar mass is just the bridge between mass and moles. this technique is actually used in real analytical chemistry. the stoichiometry problem (#11) combines gas laws with chemical reactions. start by converting everything to moles at standard conditions, then use stoichiometry to find what's produced, and finally apply gas laws to the final state. the beauty of these problems is that they're not just academic exercises - these calculations are used daily in chemical plants, environmental monitoring, and even in designing the air systems in buildings. what specific part would you like me to elaborate on? sometimes working through one problem type thoroughly helps clarify the approach for similar problems.gas laws tutorial solutions 1. derive the si of pressure pressure is defined as force per unit area: p = f/a since force = mass × acceleration (f = ma), we have: force in si units: kg⋅m⋅s⁻² area in si units: m² therefore: pressure = (kg⋅m⋅s⁻²)/m² = kg⋅m⁻¹⋅s⁻² this unit is called the pascal (pa): 1 pa = 1 kg⋅m⁻¹⋅s⁻² 2. define "pressure" using two different equations definition 1 (mechanical): p = f/a where f is the perpendicular force applied and a is the area over which it acts. definition 2 (kinetic theory): p = (1/3)nmv² where n is the number of molecules per unit volume, m is molecular mass, and v² is the mean square velocity. both equations show that the overall si unit of pressure is the pascal (pa). 3. state equation of ideal gases from empirical gas laws from the empirical laws: boyle's law: p ∝ 1/v (at constant t, n) charles's law: v ∝ t (at constant p, n) avogadro's law: v ∝ n (at constant p, t) combining these relationships: pv = nrt where r is the ideal gas constant (8.314 j⋅mol⁻¹⋅k⁻¹) 4. pressure conversions (a) 0.357 atm to torr: since 1 atm = 760 torr 0.357 atm × 760 torr/atm = 271 torr (b) 6.6 × 10⁻² torr to atmospheres: 6.6 × 10⁻² torr × (1 atm/760 torr) = 8.7 × 10⁻⁵ atm (c) 147.2 kpa to torr: first convert to atm: 147.2 kpa × (1 atm/101.325 kpa) = 1.453 atm then to torr: 1.453 atm × 760 torr/atm = 1,104 torr 5. gas constant calculation given: volume of container: 10.0 dm³ = 0.0100 m³ mass of helium: 0.126 g = 0.000126 kg temperature: 102°c = 375 k pressure: 103.20 cm h₂o at 25°c step 1: convert pressure to pa at 25°c, density of water = 0.997 g/cm³ = 997 kg/m³ p = ρgh = 997 × 9.81 × 1.0320 = 10,090 pa step 2: calculate moles of helium molar mass of he = 4.003 g/mol n = 0.126 g ÷ 4.003 g/mol = 0.0315 mol step 3: calculate r r = pv/nt = (10,090 × 0.0100)/(0.0315 × 375) = 8.55 j⋅mol⁻¹⋅k⁻¹ 6. molar mass determination given: mass: 1.007 g pressure: 715 mmhg = 0.941 atm volume: 452 cm³ = 0.452 l temperature: 23.0°c = 296 k step 1: calculate moles using ideal gas law pv = nrt n = pv/rt = (0.941 × 0.452)/(0.0821 × 296) = 0.0175 mol step 2: calculate molar mass m = mass/moles = 1.007 g/0.0175 mol = 57.5 g/mol 7. co₂ generation from limestone reaction: caco₃(s) → cao(s) + co₂(g) given: pressure: 1.3 atm temperature: 31°c = 304 k volume: 250 ml = 0.250 l using pv = nrt: n = pv/rt = (1.3 × 0.250)/(0.0821 × 304) = 0.0130 mol co₂ 8. tennis ball pressure given: volume: 144 cm³ = 0.144 l mass of n₂: 0.33 g temperature: 24°c = 297 k step 1: calculate moles of n₂ n = 0.33 g ÷ 28.01 g/mol = 0.0118 mol step 2: calculate pressure p = nrt/v = (0.0118 × 0.0821 × 297)/0.144 = 2.1 atm 9. density of carbon tetrachloride vapor given: temperature: 67 torr = 0.088 atm temperature: 4°c = 277 k for an ideal gas: pv = nrt and n = mass/m therefore: p = (mass/v) × (rt/m) = density × (rt/m) rearranging: density = pm/rt molar mass of ccl₄ = 12.01 + 4(35.45) = 153.8 g/mol density = (0.088 × 153.8)/(0.0821 × 277) = 0.59 g/l 10. argon gas pressure given: mass: 25 g volume: 1.5 dm³ = 1.5 l temperature: 30.0°c = 303 k step 1: calculate moles n = 25 g ÷ 39.95 g/mol = 0.626 mol step 2: calculate pressure if behaving as perfect gas p = nrt/v = (0.626 × 0.0821 × 303)/1.5 = 10.3 atm since the problem asks "if not, what pressure would it exert?" this suggests we should consider real gas behavior, but without additional data (like van der waals constants), we can only calculate the ideal gas pressure. 11. gas mixture reaction given: 120 cm³ nh₃ at 25°c and 750 mmhg 165 cm³ o₂ at 50°c and 635 mmhg final vessel: 0.30 dm³ at 150°c reaction: 4nh₃(g) + 5o₂(g) → 4no(g) + 6h₂o(g) this is a complex stoichiometry problem that would require:

converting all initial conditions to moles determining the limiting reagent calculating products formed finding final pressure using ideal gas law the total pressure after reaction would be calculated using: p_total = n_total × rt/v where n_total includes unreacted reactants plus all products, and t = 150°c = 423 k.

Soumya181282 · Answer

the gaseous state is one of the fundamental states of matter, alongside solid, liquid, and plasma. in this state, substances exist as gases, characterized by specific physical properties and behaviors. 🔬 definition: the gaseous state is the state of matter in which particles (atoms or molecules) are widely spaced, move rapidly, and do not have a fixed shape or volume. gases fill any container they are placed in. 🌬️ key characteristics of gases:

no fixed shape or volume: gases take the shape and volume of their container. highly compressible: particles are far apart and can be pushed closer together. low density: gases have fewer particles in a given volume compared to solids and liquids. high kinetic energy: gas molecules move rapidly in all directions. exerts pressure: gas particles collide with the walls of their container, creating pressure. diffusion and effusion: gases spread and mix rapidly due to high particle motion. 🧪 important gas laws: lawstatementmathematical formboyle's lawpressure ∝ 1/volume (at constant t)

1 pa=1 n/m21\ ext{pa} = 1\ ext{n/m}^2

charles's lawvolume ∝ temperature (at constant p)

vt=constant\frac{v}{t} = ext{constant}

gay-lussac's lawpressure ∝ temperature (at constant v)

pt=constant\frac{p}{t} = ext{constant}

avogadro’s lawvolume ∝ number of moles (at constant p, t)

d=13.56822.747≈0.597 g/ld = \frac{13.568}{22.747} \approx 0.597~ ext{g/l}

ideal gas equationcombines all laws

p v = n rt

🌡️ ideal vs. real gases: ideal gas: obeys all gas laws under all conditions. assumes no intermolecular forces and negligible volume of particles. represented by the ideal gas equation

p v = n rt

. real gas: deviates from ideal behavior at high pressure and low temperature. van der waals equation is used to describe real gas behavior:

(p+av2)(v−b)=rt\left(p + \frac{a}{v^2} ight)(v - b) = rt

where

a

and

b

are constants specific to each gas. 📚 applications of gaseous state: breathing (o₂ and co₂ exchange) balloons, airbags, and gas cylinders weather prediction (atmospheric gases) rocket propulsion and combustion particles in the gaseous state, a fundamental state of matter, are widely separated, move quickly, and lack a fixed shape or volume. gases have several important properties, including no fixed shape or volume, high compressibility, low density, high kinetic energy, exerting pressure, diffusion, and effusion. they also fill any container they are placed in. avogadro's law, charles' law, gay-lussac's law, and boyle's law are significant gas laws. while the real gas equation deviates from ideal behavior at low temperatures and high pressures, the ideal gas equation incorporates all of the laws. the behavior of real gases is described by the van der waals equation. applications for the gaseous state include breathing, airbags, balloons, gas cylinders, weather forecasting, rocket propulsion, and combustion. the ideal gas obeys

v_6q · Answer

the gaseous state is a physical state of matter where particles are widely spaced, move freely, and have neither a fixed shape nor volume. gases are highly compressible and expand to fill any container they occupy. their behavior is described by gas laws such as boyle’s law, charles’s law, and the ideal gas law (

pressure=forcearea ext{pressure} = \frac{ ext{force}}{ ext{area}}

). unlike solids and liquids, gas particles have minimal intermolecular forces and move rapidly in all directions. while ideal gases follow these laws perfectly, real gases show deviations under high pressure and low temperature due to particle interactions and finite volume.

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Price:	25.00
Added:	14 July 2025, 02:00
Words:	1025 words
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