Question

A gas sample of helium is kept at some temperature and the initial pressure of gas is $P_o=5kPa$ and initial volume is $V_o=800cc$. Now, if gas starts to expand such that the internal energy of gas remains constant and final volume becomes $1600cc$, then the work done by gas is (Take $\ln 2=0.7$)

• A. $2.4J$
• B. $3.2J$
• C. $2.8J$
• D. $4.2J$

Answer 1

The correct option is C $2.8J$

We know that for isothermal process the internal energy remains constant and work done by the gas is
$\Rightarrow W_{isothermal}=nR{T}_{o}lo{g}_e\left(\frac{V_{final}}{V_{initial}}\right)$
$=PVlo{g}_e\left(\frac{1600}{800}\right)$
$=5\times {10}^3\times 800\times {10}^{-6}\times \left(lo{g}_{e}2\right)$
$=40\times {10}^5\times {10}^{-6}\times \left(0.693\right)$
$=4\times 0.7=2.8J$