When a real gas is placed into a container at high pressure, why might the pressure of the gas be higher than what was predicted by the ideal gas law?

Because the molecules of an ideal gas are assumed to have zero volume, the volume available to them for motion is always the same as the volume of the container....

P1V1 = P2V20.88 atm × 1.50L = 2.00L × X1.32 = 2.00XX = 1.32/2.00= 0.66 atm

Temperature= 0 °C = 273kT2 = 305k∆T = 321mmHg = 0.00131579atm 725× 0.00131579 = 0.954atmV= 0.954 × 32×1= 30.53mol/L

Partial pressure =1.37atm

The behavior of real gases usually agrees with the predictions of the ideal gas equation to within 5% at normal temperatures and pressures (Normal temperature -  20oC (293.15 K, 68oF), pressure -1 atm (101.325 kN/m2, 101.325...

Answer: pV = nRT R = 8.314 L · kPa · K-1 · mol-1 81.8 · V = 0.582 · 8.314 · (273+15) V = (0.582 · 8.314 · (273+15))/81.8...

Solution:The first step is to convert temperature to Kelvins and pressure to atm:15°C + 273 = 288 K81.8 kPa × (1 atm / 101.325 kPa) = 0.8073 atmThe Ideal Gas...

Answer on Question #83513Container A holds 782 mL of ideal gas at 2.80 atm. Container B holds 174 mL of ideal gas at 4.40 atm. If the gases are allowed...

pV=nRTT = 295 KV = 0.5 L = 0.5*10-3 m3p = 1.442 bar = 144200 Pa R = 8.314 J/mol*Kn = pV/RT = (144200*0.5*10-3)/(8.314*295) = 0,029 mol.

(a) A higher P increases NX