A gas is brought to a final pressure of 6.8 atm. after increasing its temperature from 40 K to 280 K. Calculate the original pressure of the gas.

0.97 atm.

Explanation:

From the question given above, the following data were obtained:

Final pressure (P2) = 6.8 atm

Initial temperature (T1) = 40 K

Final temperature (T2) = 280 K

Initial pressure (P1) =?

Thus, we can obtain the initial (original) pressure of the gas as follow:

P1/T1 = P2/T2

P1 /40 = 6.8/280

Cross multiply

P1 × 280 = 40 × 6.8

P1 × 280 = 272

Divide both side by 280

P1 = 272/280

P1 = 0.97 atm

Therefore, the original pressure of the gas is 0.97 atm.

d: the voltage of the secondary coil is less than that of the primary coil.

im not suree sorry if wrong

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 :