Work Done by the gas in Isothermal Condition against constant external pressure

**Question** :
\*



> 
> Under isothermal condition, a gas at 300 K expands from 0.1 L to 0.25L
> against a constant external pressure of 2 bar . The work done by the
> gas is( given that 1L.bar=100J)
> 
> 
> 


* 


**Answer** (from internet, that I always got) :



> 
> W=−Pext.ΔV=−2×0.15bar=−2×0.15×100J=−30J
> 
> 
> 


I feel that in question they had asked the Work done by the gas, thus we do not have to use negative sign, as the volume is increase so, gas has done positive work. But, I am not able to find not a single solution that supported me, as it is the question and answer of NEET-2019 (A national level exam in India for Medical Students). So, please help to get the right answer, and suggest me if any correction is required within me.




Be aware you deal with 2 opposite sign conventions for work.


IUPAC and physical chemistry supports the convention Q and W have positive sign when done on the system, not by the system. It means


$$\Delta U = Q + W,$$


with the positive work and heat increasing the system internal energy.


Historically, engineering and/or physics have been more inclined toward


$$\Delta U = Q - W,$$


i.e. positive work was the work done by the system, decreasing system internal energy.


Currently, at least physics in large extent shifts toward the former convention. There are many exceptions and there is a vaque boundary of convention domains. One has always to check or write, what convention was/is used.


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