Question

A gas that has a volume of 28 liters, a temperature of 45 °C, and an unknown pressure has its volume increased

to 34 liters and its temperature decreased to 35 °C. If I measure the pressure after the change to be 2.0 atm,

what was the original pressure of the gas?

Answer 1

Answer:

2.5 atm

Explanation:

V1 = 28L

T1 = 45°C = (45 + 273.15)K = 318.15K

V2 = 34L

T2 = 35°C = (35 + 273.15)K = 308.15k

P2 = 2.0atm

P1 = ?

From general gas equation,

(P1 × V1) / T1 = (P2 × V2) / T2

P2 × V2 × T1 = P1 × V1 × T2

P1 = (P2 × V2 × T1) / (V1 × T2)

P1 = (2.0 × 34 × 318.15) / (28 × 308.15)

P1 = 21634.5 / 8628.2

P1 = 2.5 atm

The initial pressure of the gas is 2.5atm

Answer 2

The original (initial) pressure of the gas in discuss is; 2.5atm

By the ideal gas equation which relates the temperature, pressure and volume of a gas at two different states;

• (P1 V1)/T1 = (P2 V2)/T2

In which case;

• V1 = 28L

• T1 = 45°C = (45 + 273.15)K = 318.15K

• V2 = 34L

• T2 = 35°C = (35 + 273.15)K = 308.15k

• P2 = 2.0atm

• P1 = ?

According to the general gas equation,

• (P1 × V1) / T1 = (P2 × V2) / T2

• P2 × V2 × T1 = P1 × V1 × T2

• P1 = (P2 × V2 × T1) / (V1 × T2)

• P1 = (2.0 × 34 × 318.15) / (28 × 308.15)

• P1 = 21634.5 / 8628.2

P1 = 2.5 atm

Ultimately, The initial pressure, P1 of the gas is 2.5atm

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