Question

When 60.0 g methane (CH4) is placed in a 1.00-L vessel, the pressure is measured to be 130 atm. Calculate the temperature of the gas using (a) the ideal gas law and (b) the van der Waals equation. Do attractive or repulsive forces dominate

Answer 1

(a) By the ideal gas law, the temperature of the gas is approximately 423.51129 K

(b) By the Van der Waals equation, the temperature of the gas is approximately 442.00558 K

Step-by-step explanation:

The given parameters are;

The mass of the methane gas, m = 60 g

The volume of the container in which the gas is placed, V = 1.00-L

The pressure of the gas in the container, P = 130 atm

The molar mass of methane, CH₄ = 16.04 g/mol

The Van der Waals Constant for Methane (CH₄) a = 2.253 atm L²·mole⁻² and b = 0.04278 L·mol⁻¹

The universal gas constant, R = 0.08206 L·atm·mol⁻¹·K⁻¹

The number of moles of methane present, n = 60.0 g/(16.04 g/mol) ≈ 3.7406 48 moles

(a) The ideal gas law is P·V = n·R·T

Where;

P = The pressure of the gas

V = The volume of the gas

R = The universal gas constant

T = The temperature of the gas

n = The number of moles of the gas

Therefore, T = P·V/n·R

By substituting the known values of the variables, we get;

T = 130×1/(3.740648 × 0.08206) ≈ 423.51129 K

The temperature of the gas, T ≈ 423.51129 K

(b) The Van der Waals equation of state;

`n \cdot R \cdot T = \left (P + a \cdot (n^2)/(V^2) \right) \cdot (V - n\cdot b)`

Therefore, we have;

`T = ( \left (P + a \cdot (n^2)/(V^2) \right) \cdot (V - n\cdot b))/(n \cdot R)`

Therefore, we have;

`T = ( \left (130 + 2.253 * (3.740648^2)/(1^2) \right) * (1 - 3.740648\cdot 0.04278))/(3.740648 * 0.08206) = 442.00558 \ K`

The temperature of the gas, T ≈ 442.00558 K.