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.

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :