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:

$w=-p(V_2-V_1)$

$w=-(58.0atm)\times (61.0-33.0)L$

$w=-1624L.atm=-1624\times 101.3J=-164511.2J=-164.5kJ\aprrox -164kJ$

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

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