Answer:
(a)
P₂ = 7.13 atm
(b)
T₂ = 157.14 K
Explanation:
(a)
V₁ = initial volume = 3.7 L = 3.7 x 10⁻³ m³
V₂ = final volume = 0.85 L = 0.85 x 10⁻³ m³
P₁ = Initial Pressure of the gas = 0.91 atm = 0.91 x 101325 = 92205.75 Pa
P₂ = Final Pressure of the gas = ?
Using the equation
[tex]P_{1} V{_{1}}^{\gamma } = P_{2} V{_{2}}^{\gamma }[/tex]
[tex](92205.75) (3.7\times 10^{-3})^{1.4 } = P_{2} (0.85\times 10^{-3})^{1.4 }[/tex]
[tex]P_{2}[/tex] = 722860 Pa
[tex]P_{2}[/tex] = 7.13 atm
(b)
T₁ = initial temperature =283 K
T₂ = Final temperature = ?
using the equation
[tex]P{_{1}}^{1-\gamma } T{_{1}}^{\gamma } = P{_{2}}^{1-\gamma } T{_{2}}^{\gamma }[/tex]
[tex](92205.75)^{1-1.4 } (283)^{1.4 } = (722860)^{1-1.4 } T{_{2}}^{1.4 }[/tex]
T₂ = 157.14 K