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PLEASE HELP

A 700.0 mL gas sample at STP is compressed to a volume of 200.0 mL, and the temperature

is increased to 30.0°C. What is the new pressure of the gas in Pa?

(5 Points)

394 kPa

532 Pa

3.94 x 10^5 Pa

Respuesta :

Answer:

P2≈393.609Kpa so I think the answer is 394 kPa

Explanation:

PV=mRT Ideal Gas Law

m and R are constant because they dont change for the problem. That means

PV/T=mR = constant

so P1*V1/T1=P2*V2/T2 and note that the temperatures are in absolute temperatures (Kelvin) because you can't divide by zero.

So P2 = P1*V1*T2/(V2*T1) = 101325 Pa * 700 mL * 303K/(200 mL*273K)

P2 = 393609 Pa

drpelezo

Answer:

3.94 x 10⁵ pascals

Explanation:

combined gas law problem

P₁V₁/T₁ =  P₂V₂/T₂ => P₂ = P₁(V₂/V₁)(T₁2T₁)

P₁ = 1 atm at STP       P₂ = unknown

V₁ = 700 ml               V₂ = 200 ml

T₁ = 0°C = 273K        T₂ = 30°C = 303K

P₂ =  1atm(700ml/200ml)(303K/273K) = 3.89 atm

3.89 atm = 3.89 atm(1.01 x 10⁵Pa/atm) = 3.94 x 10⁵ pascals

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