Question

Question 22 A mixture of krypton and argon gas is expanded from a volume of 33.0L to a volume of 61.0L , while the pressure is held constant at 58.0atm . Calculate the work done on the gas mixture. Be sure your answer has the correct sign (positive or negative) and the correct number of significant digits.

Answer 1

Answer : The work done on the gas mixture is -164 kJ

Explanation :

Formula used :

`w=-p\Delta V\\\\w=-p(V_2-V_1)`

where,

w = work done = ?

p = pressure of the gas = 58.0 atm

`V_1` = initial volume = 33.0 L

`V_2` = final volume = 61.0 L

Now put all the given values in the above formula, we get:

`w=-p(V_2-V_1)`

`w=-(58.0atm)* (61.0-33.0)L`

`w=-1624L.atm=-1624* 101.3J=-164511.2J=-164.5kJ\aprrox -164kJ`

conversion used : (1 L.atm = 101.3 J)

Therefore, the work done on the gas mixture is -164 kJ