R ≈ 0.08206 L·atm·mol⁻¹·K⁻¹. 8.314 has units J·mol⁻¹·K⁻¹, not L·atm units.

R = 8.314 J·mol⁻¹·K⁻¹ (SI). 0.08206 is for L·atm units; 1.38×10⁻²³ is Boltzmann constant k_B.

R = N_A·k_B, where N_A is Avogadro's number (≈ 6.022×10²³ mol⁻¹).

Standard atmospheric pressure 1 atm = 101325 Pa (exact conventional value used in calculations).

8.314 J·mol⁻¹·K⁻¹ equals 8.314 kPa·L·mol⁻¹·K⁻¹ (since 1 J = 1 kPa·L). Use 8.314 when P in kPa and V in L.

At STP 1 mol gas ≈ 22.4 L, so 11.2 L ≈ 0.5 mol. (Using PV = nRT gives same result.)

When using R = 0.08206, R's units are L·atm·mol⁻¹·K⁻¹, so PV must be in atm·L to be consistent.

0.08206 L·atm → 0.08206×101325 Pa×1e−3 m³ = 8.314 Pa·m³·mol⁻¹·K⁻¹ which equals 8.314 J·mol⁻¹·K⁻¹.

n = PV/RT = (2.0×10.0)/(0.08206×300) ≈ 20 / 24.618 ≈ 0.813 mol (check units: atm·L consistent with R).

At STP 1 mol ideal gas occupies ≈ 22.414 L (commonly approximated as 22.4 L·mol⁻¹).

From n = m/M and PV = nRT → P = (m/V)·(RT/M) → density ρ = m/V = PM/RT.

ρ = PM/RT = (1 atm × 44.0 g/mol)/(0.08206×300 K) ≈ 44 / 24.618 ≈ 1.788 g·L⁻¹.

Dalton's law: partial pressure equals mole fraction times total pressure, P_i=x_iP_total.

Mole fraction x_B=3/(2+3)=3/5=0.6, so P_B=0.6×1.0 atm = 0.6 atm.

By definition, 760 mm Hg = 1 atm. Use conversion 760 mm Hg / 760 = 1 atm.

k_B ≈ 1.380649×10⁻²³ J·K⁻¹. R = N_Ak_B (N_A ≈ 6.022×10²³).

Compute: numerator = 0.250×0.08206×300 = 6.1545; divide by 5.00 → ≈1.2309 atm (approx 1.23 atm).

8.314 has SI units J·mol⁻¹·K⁻¹, equivalent to Pa·m³·mol⁻¹·K⁻¹.

Gas constant per particle is k_B. k_B = R / N_A, so both A and B are equivalent.

PV = nRT → P ∝ T for fixed n and V. If T doubles, P doubles proportionally.

T(°C) = T(K) − 273.15 → 300 − 273.15 = 26.85 °C.

1 m³ = 1000 L because 1 L = 10⁻³ m³, so the reciprocal gives 1000 L per m³.

Calculate V = RT/P = 0.08206×298.15 ≈ 24.45 L·mol⁻¹ (commonly quoted ≈ 24.5 L at 25 °C and 1 atm).

Avogadro's number N_A ≈ 6.02214076×10²³ entities per mole (defined constant in SI).

From ρ = PM/RT → rearrange M = ρRT/P (use consistent units: P in atm with R=0.08206 and ρ in g·L⁻¹).

M = ρRT/P = 1.25×0.08206×273.15 ≈ 1.25×22.414 ≈ 28.02 g·mol⁻¹ (approx nitrogen ~28 g·mol⁻¹).

Use R = 8.314 (J·mol⁻¹·K⁻¹) because 1 J = 1 Pa·m³, so PV in Pa·m³ matches units of RT.

v_rms=√(3RT/M) when M in kg·mol⁻¹ and R in J·mol⁻¹·K⁻¹. Ensure consistent units (M must be per mole mass in kg).

Use k_B for per-particle energy; use R for per-mole energy. Choose consistent level: molecule vs mole.

R's numeric value depends on unit system (e.g., 0.08206 L·atm vs 8.314 J). Use consistent units for P, V, and R to get correct answers.