Question

A gas that has a volume of 3.0990 L with a pressure of 2.50 atm at 41 °C is cooled to a temperature of 25 °C and the volume is reduced to 1850 mL, what would

be the new pressure in mmHg?
A) 3020 mmHg
B) 4.18 mmHg
C) 3.97 mmHg
D) 3017 mmHg

Answer 1

To find the new pressure in mmHg when a gas is cooled and its volume is reduced, the combined gas law is used and the calculations lead to the result of approximately 3017 mmHg, which is answer D 3017 mmHg.

Step-by-step explanation:

The mission is to find the new pressure in mmHg when a gas cools from 41 °C to 25 °C and its volume changes from 3.0990 L to 1850 mL. We can use the combined gas law, which relates the pressure, volume, and temperature of a gas.

Step 1: Convert all measurements to consistent units:

• Volume initial (V1) = 3.0990 L
• Volume final (V2) = 1850 mL = 1.850 L
• Pressure initial (P1) = 2.50 atm
• Pressure final (P2) = ? mmHg
• Temperature initial (T1) = 41 °C = 314 K (Convert °C to K by adding 273)
• Temperature final (T2) = 25 °C = 298 K

Step 2: Convert initial pressure from atm to mmHg:

• P1 in mmHg = 2.50 atm × 760 mmHg/atm = 1900 mmHg

Step 3: Apply the combined gas law:

P1 × V1 / T1 = P2 × V2 / T2

Solving for P2:

P2 = (P1 × V1 × T2) / (T1 × V2)

P2 = (1900 mmHg × 3.0990 L × 298 K) / (314 K × 1.850 L)

Step 4: Calculate the new pressure (P2) and match to the closest answer choice:

P2 = (1900 mmHg × 3.0990 L × 298 K) / (314 K × 1.850 L) ≈ 3017 mmHg

The correct answer is D) 3017 mmHg.