Answer:
See attached
Step-by-step explanation:
The empirical gas laws relate the effect of one gas-law variable upon another. The gas-law variables include Pressure (P), Volume(V), Temperature(T) and mass (n). When two of the four variables are related holding the other two constant, the relationships are called 'The Empirical Gas Laws'.
These include
Boyles Law* => P ∝ 1/V; Temp & mass remain constant => P₁·V₁ = P₂V₂
(NOTE => Boyles Law is the only 'inverse' empirical relationship)
Charles Law => V ∝ T; Press & mass remain constant => V₁/T₁ = V₂/T₂
Gay-Lussac Law => P ∝ T; Vol & mass remain constant => P₁/T₁ = P₂/T₂
Avogadro's Law => n ∝ V; Press & Temp remain constant => n₁/V₁ = n₂/V₂
Combined Law => All variables in play => P₁V₁/n₁T₁ = P₂V₂/n₂T₂
*Note => Convert all Temperature values to Kelvin when working with Empirical Gas Laws. If a different dimension is needed, convert after working with Kelvin values. => ( K = °C + 273 )
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Standards of the Gas Laws at 0°C(273K) STP
(Std Intn'l System) Other common equivalent unit values
Pressure 1 Atm 760mmHg, 101.325Kpa, 760 Torr, 14.2 lbs/in²
Volume 1 Liter
mass 1 mole
Temperature 0°C(=273K) K = °C + 273
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4. ?Atm = 758.7mmHg
758.8mmHg = 758.7mmHg x (1Atm/760mmHg) = 0.9983 Atm
5. ?Kpa = 760.00mmHg x (1 Atm/760.00mmHg) x 101.325Kpa/1Atm)
= 101.325Kpa
6. Boyles Law => Decrease Pressure => Increase Volume (of gas) in bag.
P₁ = 101.325Kpa => P₂ = 99.82Kpa
P₁V₁ = P₂V₂ => V₂ = P₁V₁/P₂ = V₁(101.325Kpa/99.82Kpa) = 1.015·V₁ => That is, the volume of gas in bag will expand to 1.015 x that at sea level. The bag may burst if not elastic.
7. ?Kpa = 2.56Atm
2.56Atm = 2.56Atm x 101.325Kpa/1Atm = 259.392Kpa