Respuesta :
Answer:
(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
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;
[tex]n \cdot R \cdot T = \left (P + a \cdot \dfrac{n^2}{V^2} \right) \cdot (V - n\cdot b)[/tex]
Therefore, we have;
[tex]T = \dfrac{ \left (P + a \cdot \dfrac{n^2}{V^2} \right) \cdot (V - n\cdot b)}{n \cdot R}[/tex]
Therefore, we have;
[tex]T = \dfrac{ \left (130 + 2.253 \times \dfrac{3.740648^2}{1^2} \right) \times (1 - 3.740648\cdot 0.04278)}{3.740648 \times 0.08206} = 442.00558 \ K[/tex]
The temperature of the gas, T ≈ 442.00558 K.
