Question

Suppose that a certain quantity of methane occupies a volume of 0.138 L under a pressure of 300 atm at 200 °C, and the volume required at 600 atm at 0 °C. For 300 atm and at 200 °C, Z=1.067, while for 600 atm at 0 °C, Z=1.367.​

Answer 1

Therefore, the volume required at 600 atm and 0 °C is 0.319 L.

Step-by-step explanation:

We can use the ideal gas law to solve for the number of moles of methane present, assuming ideal gas behavior at both conditions:

PV = nRT

At 300 atm and 200 °C:

n = PV/RT = (300 atm * 0.138 L) / [(0.08206 L atm mol^-1 K^-1) * (200 + 273.15) K * 1.067]

n = 2.451 mol

At 600 atm and 0 °C:

n = PV/RT = (600 atm * V2) / [(0.08206 L atm mol^-1 K^-1) * (273.15 K) * 1.367]

n = 7.682 V2

Since the number of moles of methane must be the same at both conditions:

2.451 mol = 7.682 V2

Solving for V2:

V2 = 0.319 L

Therefore, the volume required at 600 atm and 0 °C is 0.319 L.