An ideal gas is kept in a cylinder of volume $3 \mathrm{~m}^3$ at a pressure of $3 \times 10^5 \mathrm{~Pa}$…
An ideal gas is kept in a cylinder of volume $3 \mathrm{~m}^3$ at a pressure of $3 \times 10^5 \mathrm{~Pa}$. The energy of the gas is
$13.5 \times 10^6 \mathrm{~J}$
$1.35 \times 10^5 \mathrm{~J}$
$13.5 \times 10^5 \mathrm{~J}$
$135 \times 10^6 \mathrm{~J}$
Solution
For ideal gas, $V=3 \mathrm{~m}^3, \mathrm{P}=3 \times 10^5 \mathrm{pa}$
$\therefore \quad$ The energy of the gas is
$\mathrm{E}=\frac{3}{2} \mathrm{pV}=\frac{3}{2} \times 3 \times 10^5 \times 3=13.5 \times 10^5 \mathrm{~J}$