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

Answer:

Therefore, the new pressure is 417.455 kPa

Step-by-step explanation:

As per general gas equation

$\frac{P_{1}V_{1}}{T_{1} }=\frac{P_{2}V_{2}}{T_{2} }$

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

$\frac{P_{1}}{T_{1} }=\frac{P_{2}}{T_{2} }$

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}

$P_{2}=\frac{(748)(293)}{525}$

$P_{2}= 417.455$ kPa

Therefore, the new pressure is 417.455 kPa