Agas in a cylinder is held at a constant pressure of 1.80 * 105 pa and is cooled and compressed from 1.70 m3 to 1.20 m3 . the internal energy of the gas decreases by 1.40 * 105 j.
(a) find the work done by the gas.
(b) find the absolute value |q| of the heat flow into or out of the gas, and state the direction of the heat flow.
(c) does it matter whether the gas is ideal? why or why not

This question is wrong because of not correct values.The Correct question is

A gas in a cylinder is held at a constant pressure of 1.80×10⁵Pa and is cooled and compressed from 1.70m³ to 1.20m³. The internal energy of the gas decreases by 1.40×10⁵J.

(a) Find the work done by the gas.

(b) Find the absolute value of the heat flow, |Q|, into or out of the gas, and state the direction of the heat flow.

(c) Does it matter whether the gas is ideal? Why or why not?

(a) W= -9×10⁴J

(b) |Q|=2.3×10⁵

(c) It does not matter whether gas is ideal or non-ideal

Explanation:

Given

V₁=1.70m³

V₂=1.20m³

p=1.8×10⁵ pa

ΔU= -1.40×10⁵J

For (a) work done by gas

For (b) Heat flow |Q|

|Q|=ΔU+W

For (c) part

It does not matter whether the gas is ideal or not because the first law of thermodynamics which applied in our solution could applied to any material ideal or non ideal  

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s