a cylinder of helium gas has a volume of 1.0 L. The gas in the cylinder exerts a pressure of 800 kPa at 30 degrees. What volume would this gas occupy at SATP

Md Ariful Islam

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SATP - Standard Ambient Temperature and Pressure.

SATP: T = 25∘C and P = 1 bar.

P1 = 800 kPa;

V1 = 1.0 L;

T1 = 30∘C = 303 K;

P2 = 1 bar = 100 kPa;

V2 = unknown;

T2 = 25∘C = 298 K.

Solution:

The Combined Gas Law can be expressed as:

P1V1 / T1 = P2V2 / T2

Cross-multiply to clear the fractions:

P1V1T2 = P2V2T1

Divide to isolate V2:

V2 = (P1V1T2) / (P2T1)

Plug in the numbers and solve for V2:

V2 = (800 kPa × 1.0 L × 298 K) / (100 kPa × 303 K) = 7.868 L

V2 = 7.87 L

Answer: The volume of gas will be 7.87 L at SATP.

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