Question

A sample of gas is compressed by an average pressure of $0 .50$ atmosphere so as to decrease its volume from $400 \, cm^{3}$ to $200\, cm^{3}$ . During the process $8 .00 \,J$ of heat flows out to surroundings. The change in internal energy of the system is

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

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 $q=-8.00 \, J$
$\Rightarrow \text{ΔE} = - 8 + 10.13 = 2.13 \, \text{Joule}$