Question

A gas in a cylinder held at a constant pressure of $1.7\times {10}^{5}Pa$ and is cooled and compressed from $1.20m^{3}to0.8m^3$. The internal energy of the gas decreses by $1.1\times {10}^{5}J$. The magnitude of the heat flow into or out of the gas is given as $x\times {10}^4$ J. Find $x$ ($x$ is a two digit integer)

Answer 1

Work done by a gas is given by:
$w=\int PdV=P(V_2-V_1)=1.7\times {10}^5(0.8-1.2)=-0.68\times {10}^{5}J$
Given that internal energy decreases by $\Delta U=-1.1\times {10}^{5}J$
The first law of thermodynamics gives us the relation:
$q=\Delta U+w$
$q=-1.1\times {10}^5-0.68\times {10}^5=-1.78\times {10}^{5}J$
$q=-1.78\times {10}^5=-178\times {10}^{3}J$
$x=178$
If it was $x\times {10}^4$ then $x$ would be $18$.