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Gas Law Problems 1. 22.4 liters of a gas is originally at STP. calculate the indicated quantity when the conditions are changed. a. Calculate the volume is the temperature increases to 765 K at constant pressure. b. Calculate the pressure if the volume is compressed to 250.mL at constant temperature. Pi c. Calculate the temperature if the volume drops to 16.2 L and the pressure is reduced to 200. Torr.

User Zankhna
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Answer:

Step-by-step explanation:

To solve the given gas law problems, we can use the ideal gas law equation:

PV = nRT

where:

P = pressure (in atmospheres or Torr)

V = volume (in liters)

n = number of moles

R = ideal gas constant (0.0821 L·atm/(mol·K) or 62.36 L·Torr/(mol·K))

T = temperature (in Kelvin)

Given information:

Initial volume (V1) = 22.4 L

Initial pressure (P1) = 1 atm (at STP)

Final temperature (T2) = 765 K (part a)

Final volume (V2) = 250 mL = 0.25 L (part b)

Final pressure (P2) = Unknown (part b)

Final volume (V2) = 16.2 L (part c)

Final pressure (P2) = 200 Torr (part c)

a) Calculate the volume when the temperature increases to 765 K at constant pressure:

Using the equation PV = nRT, we can write:

P1V1/T1 = P2V2/T2

Plugging in the known values, we get:

(1 atm) * (22.4 L) / (273 K) = P2 * V2 / (765 K)

Solving for V2, we find:

V2 = (1 atm) * (22.4 L) * (765 K) / (273 K)

V2 ≈ 63.38 L

Therefore, the volume when the temperature increases to 765 K at constant pressure is approximately 63.38 L.

b) Calculate the pressure if the volume is compressed to 250 mL at constant temperature:

Using the equation PV = nRT, we can write:

P1V1/T1 = P2V2/T2

Plugging in the known values, we get:

(1 atm) * (22.4 L) / (273 K) = P2 * (0.25 L) / (273 K)

Solving for P2, we find:

P2 = (1 atm) * (0.25 L) / (22.4 L)

P2 ≈ 0.0112 atm

Therefore, the pressure when the volume is compressed to 250 mL at constant temperature is approximately 0.0112 atm.

c) Calculate the temperature if the volume drops to 16.2 L and the pressure is reduced to 200 Torr:

Using the equation PV = nRT, we can write:

P1V1/T1 = P2V2/T2

Plugging in the known values, we get:

(1 atm) * (22.4 L) / (273 K) = (200 Torr) * (16.2 L) / T2

Solving for T2, we find:

T2 = (200 Torr) * (16.2 L) * (273 K) / (1 atm) * (22.4 L)

T2 ≈ 263.03 K

Therefore, the temperature when the volume drops to 16.2 L and the pressure is reduced to 200 Torr is approximately 263.03 K.

User Stereoscott
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