37.2 liters of a gas has a pressure of 362.43 kPa at 46.5 °C. If the pressure increases to 693.9 kPa and the temperature to 149.2 °C, what would be the new volume of the gas?
Select one:
Oa. 25.7
Ob. 62.3
Oc. 54
O d. 25.672

Option A. 25.7 L.

Explanation:

Step 1:

Data obtained from the question. This include the following:

Initial volume (V1) = 37.2 L

Initial pressure (P1) = 362.43 kPa

Initial temperature (T1) = 46.5 °C

Final pressure (P2) = 693.9 kPa

Final temperature (T2) = 149.2 °C

Final volume (V2) =.?

Step 2:

Conversion of celsius temperature to Kelvin temperature.

This can be achieved as shown below:

Temperature (K) = Temperature (°C) + 273

T (K) = T (°C) + 273

Initial temperature (T1) = 46.5 °C

Initial temperature (T1) = 46.5 °C + 273 = 319.5 K.

Final temperature (T2) = 149.2 °C

Final temperature (T2) = 149.2 °C + 273 = 422.2 K.

Step 3:

Determination of the new volume.

This can be obtained by using the general gas equation as shown below:

Initial volume (V1) = 37.2 L

Initial pressure (P1) = 362.43 kPa

Initial temperature (T1) = 319.5 K

Final pressure (P2) = 693.9 kPa

Final temperature (T2) = 422.2 K

Final volume (V2) =.?

P1V1/T1 = P2V2/T2

362.43×37.2/319.5 = 693.9 × V2/422.2

Cross multiply

319.5×693.9×V2 = 362.43×37.2×422.2

Divide both side by 319.5 × 693.9

V2 = (362.43×37.2×422.2)/(319.5×693.9)

V2 = 25.7 L

Therefore, the new volume is 25.7 L.

- lead + phosphoric acid = hydrogen + lead(ii) phosphate

58750.08068 is the answer i think