Answer :
Answer:
45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.
Explanation:
Using Ideal gas equation for same mole of gas as
[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]
Given ,
V₁ = 25.0 L
V₂ = ?
P₁ = 2575 mm Hg
Also, P (atm) = P (mm Hg) / 760
P₁ = 2575 / 760 atm = 3.39 atm
P₂ = 1.35 atm
T₁ = 353 K
T₂ = 253 K
Using above equation as:
[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]
[tex] \frac{{3.39}\times {25.0}}{353}=\frac{{1.35}\times {V_2}}{253}[/tex]
[tex]\frac{1.35V_2}{253}=\frac{3.39\times \:25}{353}[/tex]
Solving for V₂ , we get:
V₂ = 45.0 L
45.0 L is the volume of gas will the balloon contain at 1.35 atm and 253 K.