How much work (in J) is involved in a chemical reaction if the volume decreases from 33.6 L to 11.2 L against a constant pressure of 90.5 kPa?

Video transcript

Hi everyone here we have a question telling us that initially a balloon has a volume of 14.8 L. It was released outside and expands to a volume of 18.6 L at 85.4 kill a pascal's external pressure. And our goal here is to calculate the work for this process. So some things we need to know is one atmosphere times later Equals 0.3 jewels. and one kill pascal Equals 10 to the 3rd pascal's. So let's convert our kill pascal's two atmospheres. So we have 85 0. killer pascal. We're going to multiply that by 10 to the third pascal's over one kill pascal and we're going to multiply that by one atmosphere over 101,000 Pascal's. So our killer pass skills are canceling out and our past skills are canceling out leaving us with atmospheres And that equals 0.8428 atmospheres. And now we're going to use the formula works equals negative pressure times are changed in volume. So our work is going to equal 0.84 two eight atmosphere times 18.6 l -14.8 L. So that is going to equal zero 8428 atmosphere times three 0. zero leaders. So that equals negative 3. 02 eight leaders. And now we're going to change that to jules. So we have our negative three point 2028 leaders times atmospheres. We're going to multiply that by 101 0. jewels over one atmosphere times later and our Leader Times Atmosphere are going to cancel out And that equals negative. 324 point for jules. And the negative sign means that the work was done by the system so Negative. jewels of work was done by the system. Thank you for watching. Bye.