Question

A sample of gas is compressed by an average pressure of $0.50$ atmosphere so as to decrease its volume from $400c{m}^3$ to $200c{m}^3$ . During the process $8.00J$ of heat flows out to surroundings. The change in internal energy of the system is

• A. $+2.13J$
• B. $+10.13J$
• C. $-2.13J$
• D. $-10.13J$

Answer 1

Answer: (A)

Here, $\Delta V=200-400=-200\text{c}{\text{m}}^3$
As we know $1\text{Litre}=1000{\text{ cm}}^{\text{3}}$
$\Rightarrow \text{ΔV}=-0.2\text{Litre}$ .
External Pressure $\left(P\right)=0.50$ then
Work done $=-P\Delta V=+0.50\left(0.2\right)$
$=+0.1\text{atm}\text{Litre}$
and $\text{1}\text{litre-atmosphere}\text{=}\text{101}\text{.3}\text{Joule}$
$\therefore W=+10.13$ Joule
According to First law of thermodynamics
$q=\Delta E-W$
then $\Delta E=q+W$
Given $q=-8.00J$
$\Rightarrow \text{ΔE}=-8+10.13=2.13\text{Joule}$