Question

A 22.4 L sample of argon gas at 101.3 kPa and 0°C is cooled to -15°C and

the volume decreases to 19.7 L. What is the new pressure of the gas in
KPa? Remember temperature should be in K!

Answer 1

The new pressure of the argon gas, after cooling from 0°C to -15°C and a volume change from 22.4 L to 19.7 L, is 113.89 kPa.

Step-by-step explanation:

We are given a sample of argon gas at an initial condition of 101.3 kPa and 0°C, with a volume of 22.4 L. To find the new pressure after the gas is cooled to -15°C with a volume decrease to 19.7 L, we can use the combined gas law:

P1V1/T1 = P2V2/T2

First, we convert temperatures to Kelvin:

• T1 = 0°C + 273.15 = 273.15 K
• T2 = -15°C + 273.15 = 258.15 K

Now, we can plug in the values we know to solve for P2 (the new pressure).

P1 = 101.3 kPa

V1 = 22.4 L

T1 = 273.15 K

V2 = 19.7 L

T2 = 258.15 K

Rearranging the formula to solve for P2 gives us:

P2 = (P1V1T2) / (T1V2)

P2 = (101.3 kPa × 22.4 L × 258.15 K) / (273.15 K × 19.7 L)

By calculating, we find:

P2 = (101.3 kPa × 22.4 L × 258.15 K) / (273.15 K × 19.7 L) = 113.89 kPa

Therefore, the new pressure of the argon gas is 113.89 kPa.