91Ó°ÊÓ

Give,

The gas is compressed from$V_1=1.7m^3$ to$V_2=1.2m^3$ and a constant pressure $p=1.8x{10}^5\mathrm{Pa}$, whereas the internal energy is decreased by$4×{10}^{5}J$

Since the pressure is kept constant, hence the work done is given by:

$\begin{array}{ll}W=p\left(V_2+V_1\right)& \\ =1.8×{10}^{5}Pa×\left(1.2m^3-1.7m^3\right)& \\ =-9×{10}^{4}J& \end{array}$

Therefore, the work done is$-9×{10}^{4}J$ , since the work done is negative work is done on the gas by the surrounding.

Now, according to the first law of thermodynamics the heat flow is given by:

$Q=∆U+W$

Since the internal energy is decreased that means the change of energy is negative $(∆U=-1.4×{10}^{5}J)$and we have also proved that the work done is negative. Now using the first law of thermodynamics to calculate the heat flow through the gas, we have:

$\begin{array}{l}Q=∆U+W\\ =-1.4×{10}^{5}J+\left(-9×{10}^{4}J\right)\\ =-2.3×{10}^{5}J\end{array}$

Now the absolute value of would be

$$\begin{gathered} \left|Q\right|=\left|-2.3×{10}^{5}J\right| \\ =2.3×{10}^{5}J \end{gathered}$$

Therefore, the absolute value of heat flow is$2.3×{10}^{5}J$ and the direction is out from the gas to the surrounding.