Question

After 273 m3 of ethylene oxide is formed at 748 kPa and 525 K, the gas is cooled at constant volume to 293 K.The new pressure is _____kPa

Answer 1

P1/T1 = P2/T2

748 / 525 = P2/293

P2 = 748/525 X 293 = 417.455 kPa

Answer 2

Therefore, the new pressure is 417.455 kPa

Explanation:

As per general gas equation

`(P_(1)V_(1))/(T_(1) )=(P_(2)V_(2))/(T_(2) )`

If Volume of a gas is constant then equation is represented by

`(P_(1))/(T_(1) )=(P_(2))/(T_(2) )`

Here
`P_(1)=748kPa` and
`T_(1)=525K`

If
`T_(2)=293K` then we have to calculate the value of
`P_(2)`

Now we plug in the values in the equation

\frac{748}{525}=\frac{P_{2}}{293}

`P_(2)=((748)(293))/(525)`

`P_(2)= 417.455` kPa

Therefore, the new pressure is 417.455 kPa