Answer to Question #134635 in Mechanics | Relativity for Jessica

Question

An ideal gas at 10.3 °C and a pressure of 2.48 x 10^5 Pa occupies a volume of 3.12 m3. (a) How many moles of gas are present? (b) If the volume is raised to 5.45 m3 and the temperature raised to 25.6 °C, what will be the pressure of the gas?

Answer 1

V₁ = 3.12 m³ = 3120 L

T₁ = 10.3 + 273.15 = 283.45 K

P₁ = 2.48×10⁵ Pa = 2.44 atm

V₂ = 5.45 m³ = 5450 L

T₂ = 25.6 + 273.15 = 298.75 K

P₂ = ?

(a) Ideal gas law:

PV=nRT

Molar gas constant

R=0.082058 L×atm/(K×mol)

n = PV/RT

$n = \frac{2.44\times3120}{0.082058\times283.45} = 327.3 \;mol$

(b) $\frac{P_1×V_1}{T_1} = \frac{P_2×V_2}{T_2}$

$\frac{2.44\times 3120}{283.45} = \frac{P_2\times5450}{298.75}$

26.85 = 18.24P₂

P₂ = 1.472 atm = 1.49×10⁵ Pa

Answer: 327.3 mol, 1.49×10⁵ Pa