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?

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