cylinder with initial volume of 1 L

Were you given the initial temp of the 1 l volume?  

no only this question 

Hmmm...I think I am stuck...

   if you add 5kj (or subtract) from a volume...it will change ...according to the gas laws...but if it is 6000 degrees it will change less than if it was 200 degrees.....so I think I need an initial temp.......I could be wrong....was STP stated?

Solution:

$\small \text {Translate Joules to Liter-atmospheres}\\ w = -5000J * \dfrac{ 1L * atm} {101.325J} =-49.35L*atm \,| \small \text { Negative (-) means energy entered system.}\\ w = −P*\Delta V\\ \small \text {Change in volume }\\ \Delta V = -\dfrac {w}{P} = -\dfrac {-49.35L*atm }{1 atm}=+ 49.35L \,| \small \text { Positive (+) means gas expanded.}\\ \text { }\,\\ \text {Final Volume of the cylinder } 49.35L$

GA