How would you calculate the work associated with the compression of a gas from 121.0 L to 80.0 L at a pressure of 13.1 atm?

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Answer 1 · 2017-09-30T23:54:40

$W_"on gas"~~1.75xx10^5" J"$

Assuming this is an ideal gas:

$W_"env"=-int_(v_i)^(v_f)" P dV"$

$=>W_"env"=-PDeltaV=P(V_"f"-V_"i")$

Let's convert liters to cubic meters first so that our units work out.

$=>V_"i"=0.212" m"^3$

$=>V_"f"=0.08" m"^3$

We also know that $P=13.1" atm"$ , which is $1327358 " N"//"m"^2$ .

$=>W_"env"=-(1327358 " N"//"m"^2*(0.08" m"^3-0.212" m"^3))$

$=>=-(1327358 " N"//"m"^2*-0.132" m"^3)$

$=>~~1.75xx10^5" Nm"$

$=>~~color(blue)(1.75xx10^5" J")$

$:.$ The work done on the gas by the environment to compress the gas is $~~1.75xx10^5" J"$