Suatu gas 0.25 mol menempati ruang 5 liter pada tekanan 2x10^5 Pa.Bila gas tersebut dianggap gas ideal,dengan tetapan gas universal 8.31 J.mol/K,maka temperatur

TKG
• persamaan keadaan

P = 2×10^5 Pa
V = 5 L = 5×10^-3 m³
n = 0,25 mol
T = __?

PV = n R T
T = P V / (n R)
T = 2×10^5 • 5×10^-3 / (0,25 • 8,31)
T = 10³ / 2,0775
T ≈ 481 K ← jwb

[tex]\displaystyle n=0,25\,\text{mol}\\V=5\cdot10^{-3}\,\text{m}\\p=2\cdot10^5\,\text{Pa}\\R=8,31\,\text{Pam}^3/\text{mol K}\\\\T=?\\\\pV=nRT\\2\cdot10^5\cdot5\cdot10^{-3}=\frac14\cdot8,31T\\10^3=\frac14\cdot8,31T\\\frac{4\cdot10^3}{8,31}=T\\\boxed{\boxed{481,3\,\text{K}=T}}[/tex]