Question

8) A gas that has a volume of 28 liters, a temperature of 65 0C, and an unknown pressure has its volume increased to 36 liters and its temperature decreased to 35 0C. If I measure the pressure after the change to be 2.0 atm, what was the original pressure of the gas?

Answer 1

Answer:

P₁  = 2.82 atm

Explanation:

Given data:

Initial volume = 28 L

Initial pressure = ?

Initial temperature = 65 °C (65 +273.15 = 338.15 K)

Final temperature =  35 °C (35 +273.15 = 308.15 K)

Final volume = 36 L

Final pressure = 2.0 atm

Formula:

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₁V₁/T₁ = P₂V₂/T₂

P₁ =  P₂V₂T₁/T₂ V₁  

P₁  = 2.0 atm × 36 L × 338.15 K / 308.15 K × 28 L

P₁  = 24346.8 atm .L. K / 8628.2 K.L

P₁  = 2.82 atm

Answer 2

The original pressure of the gas whose pressure changed to 2 atm is 2.82atm.

HOW TO CALCULATE PRESSURE:

• The pressure of a gas can be calculated by using the following expression:

P1V1/T1 = P2V2/T2

Where;

1. P1 = initial pressure
2. P2 = final pressure
3. V1 = initial volume
4. V2 = final volume
5. T1 = initial temperature
6. T2 = final temperature

• According to this question; P1 = ?, P2 = 2 atm, V1 = 28L, V2 = 36L, T1 = 65°C (338K), T2 = 35°C (308K)

• P1 × 28/338 = 2 × 36/308

• 8624P1 = 24336

• P1 = 24336 ÷ 8624

• P1 = 2.82atm

• Therefore, the original pressure of the gas whose pressure changed to 2 atm is 2.82atm.

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