An aluminum can holds 350 mL of gas at 0 C and 1.0 atm. what is the new volume if the can is heated to 10 C and the pressure inside the can rises to 75 atm?

4.84mL

Explanation:

Step 1:

Data obtained from the question. This include the following:

Initial volume (V1) = 350mL

Initial temperature (T1) = 0°C

Initial pressure (P1) = 1atm

Final temperature (T2) = 10°C

Final pressure (P2) = 75atm

Final volume (V2) = ... ?

Step 2:

Conversion of celsius temperature to Kelvin temperature.

This can be achieved by using the formula below:

T (K) = T (°C) + 273

Initial temperature (T1) = 0°C

Initial temperature (T1) = 0°C + 273 = 273K

Final temperature (T2) = 10°C

Final temperature (T2) = 10°C + 273 = 283K

Step 3:

Determination of the new volume.

The new volume can be obtained by using the general gas equation as follow:

P1V1/T1 = P2V2/T2

1 x 350/273 = 75 x V2/283

Cross multiply

273 x 75 x V2 = 1 x 350 x 283

Divide both side by 273 x 75

V2 = (1 x 350 x 283) / (273 x 75)

V2 = 4.84mL

Therefore, the new volume is 4.84mL

Final volume of the gas is 4.837mL

Explanation:

Initial volume (V1) = 350mL = 0.35L

Initial temperature (T1) = 0°C = (0 + 273.15) k = 273.15k

Initial pressure (P1) = 1.0atm

Final volume (V2) = ?

Final temperature (T2) = 10°C = (10 + 273.15) k = 283.15K

Final pressure (P2) = 75atm

To solve this question, we'll have to use combined gas equation which is the combination of all gas law I. e Charle's laws, Boyle's law, Pressure law etc.

According to combined gas equation,

(P1 * V1) / T1 = (P2 * V2) / T2

Make V2 the subject of formula,

V2 = (P1 * V1 * T2) / (P2 * T1)

V2 = (1.0 * 0.35 * 283.15) / (75 * 273.15)

V2 = 99.1025 / 20,486.25

V2 = 0.004837L

V2 = 4.837mL

The final volume of the gas is 4.837mL