For this problem, there are 4 things required to receive full credit. 1) The name of the gas law you are using to solve the problem. 2) The formula of the gas law you are using to solve the problem. 3) The values given in the problem set up in the appropriate places with units. 4) The answer rounded to the correct number of sig figs with units. If the pressure and temperature of 5.40 L of a gas at STP are changed to 265.5 KPa and -93.2°C, what is the new volume of the gas in milliliters?

The answer to your question is given below.

Explanation:

1. The name of the gas law is:

General gas law.

2. The formula for the gas law is:

P1V1 /T1 = P2V2 /T2

P1 is initial pressure.

P2 is final pressure.

V1 is the initial volume.

V2 is the final volume.

T1 is the initial temperature in Kelvin.

T2 is the final temperature in Kelvin.

3. Data obtained from the question.

Initial pressure (P1) = stp = 101.325KPa

Initial temperature (T1) = stp = 273K

Initial volume (V1) = 5.40L

Final pressure (P2) = 265.5 KPa

Final temperature (T2) = –93.2°C =

–93.2°C + 273 = 179.8K

4. Determination of the new volume. This is illustrated below:

P1V1 /T1 = P2V2/T2

101.325x5.4/273 = 265.5xV2/179.8

Cross multiply to express in linear form

V2x273x265.5 = 101.325x5.4x179.8

Divide both side by 273 x 265.5

V2 = (101.325x5.4x179.8)/(273x265.5)

V2 = 1.357L.

Now, we shall convert 1.357L to mL as shown below:

1L = 1000mL

Therefore, 1.357L = 1.357 x 1000 = 1357mL.

Therefore, the new volume of the gas is 1357mL

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.