Question

The internal energy of a monoatomic ideal gas is $1.5nRT$. One mole of helium is kept in a cylinder of cross-section $8.5{\text{cm}}^2$. The cylinder is closed by a light frictionless piston. The gas is heated slowly in a process during which a total of $42\text{J}$ heat is given to the gas. The temperature rises through $2^{\circ }C$. The distance moved by the piston is given as $a\times {10}^b\text{m}$ in scientific notation. Find the value of $a+b$. Here, $a,b$ are integers [Take $R=\frac{25}{3}$ in $SIunit$,atmospheric pressure =$100\text{kPa}$]

Answer 1

$1.00$
The change in internal energy of the gas is
$△U=1.5nR\left(△T\right)$
=$1.5\left(1\text{mol}\right)\frac{25}{3}\frac{J}{\text{mol-K}}\left(2\text{K}\right)$=$25\text{J}$.
The heat given to the gas = $42\text{J}$
The work done by the gas is $△W=△U=42\text{J}-25\text{J}=17\text{J}$
If the distance moved by the piston is $X$ , the work done is
$△W=\left(100\text{kPa}\right)\left(8.5{\text{cm}}^2\right)X.$
Thus,$\left(\frac{{10}^5\text{N}}{{\text{m}}^2}\right)(8.5\times {10}^{-4}{\text{m}}^2)X=17\text{J}$

$X=0.2\text{m}=20\text{cm}.$
but the given distance is
$a\times {10}^b=0.2\text{m}=2\times {10}^{-1}\text{cm}$
So, $a+b=2-1=1$