Round to 2 significant numbers, 5239

See attached

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 )

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  

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