Question

I NEEED PHYSICS HELPPPPPPPPPPP!!!!!!! :((((((( PLS

Answer 1

12) The gas compresses, and the piston lowers

13) The work done by the gas is $7.8\cdot 10^5 J$

Explanation:

12)

For a gas kept at constant pressure, the work done on the gas is given by:

$W=-p\Delta V = -p(V_f - V_i)$

where

p is the pressure of the gas

$V_f$ is the final volume of the gas

$V_i$ is the initial volume of the gas

We see that when work is done ON the gas, the work done is positive: therefore, this means that the term

$\Delta V = V_f - V_i$

must be negative. This implies that

$V_f < V_i$

So, the gas has being compressed (by the piston, which lowers down), and its final volume is smaller than the initial volume.

13)

Here we have a gas kept at a constant pressure of

$p=1.3\cdot 10^5 Pa$

And its initial volume is

$V_i = 6.0 m^3$

The gas is expanded to twice its initial volume, so the final volume is

$V_f = 2 V_i = 12.0 m^3$

The work done BY the gas is

$W=p\Delta V = p(V_f - V_i)$

Where here we removed the negative sign because this is the work done BY the gas. Substituting, we find

$W=(1.3\cdot 10^5)(12.0-6.0)=7.8\cdot 10^5 J$

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