Question

The pressure of a 70.0L sample of gas is 600 mm Hg at 20.0C. If the temperature drops to 15.0C and the volume expands to 90.0L, what will the pressure of the gas be?

Answer 1

Answer:

458.7 mmHg

Explanation:

Step 1:

Data obtained from the question. This includes:

Initial volume (V1) = 70L

Initial pressure (P1) = 600 mmHg

Initial temperature (T1) = 20°C

Final temperature (T2) = 15°C

Final volume (V2) = 90L

Final pressure (P2) =...?

Step 2:

Conversion of celsius temperature to Kelvin temperature.

This is illustrated below:

T(K) = T (°C) + 273

Initial temperature (T1) = 20°C

Initial temperature (T1) = 20°C + 273 = 293K

Final temperature (T2) = 15°C

Final temperature (T2) = 15°C + 273 = 288K

Step 3:

Determination of the new pressure of the gas.

The new pressure of the gas can be obtained by using the general gas equation as shown below:

P1V1/T1 = P2V2/T2

600 x 70/293 = P2 x 90/288

Cross multiply to express in linear form

P2 x 90 x 293 = 600 x 70 x 288

Divide both side by 90 x 293

P2 = (600 x 70 x 288) / (90 x 293)

P2 = 458.7 mmHg

Therefore, the new pressure of the gas is 458.7 mmHg