Question

A cylinder of ideal gas is closed by a $8\;Kg$ movable piston of area $60cm^{2}$. The atmospheric pressure is $100\;kPa$. When the gas is heated from $30^{\circ}C$ to $100^{\circ}C$ the piston rises $20\;cm$. The piston is then fastened in the place and the gas is cooled back to $30^{\circ}C$. If $\Delta Q_{1}$ is the gas during heating and $\Delta Q_{2}$ is the heat lost during cooling, then the difference is $(130+x)$. Find the value of $x$.

Answer 1

Hint: We know from the first law of thermodynamics, that the energy of an isolated system is conserved. We can calculate the change in heat and the change in internal energy from the formula given below.
Formula: $\Delta Q=\Delta U+\Delta W$, $\Delta U=nC\Delta T$

Complete answer:
Ideal gas law or the general gas equation is the combination of Boyles Law, Charles’s law, Avogadro’s law and Gay Lussac’s law. It gives the relationship between the pressure $P$ applied on a $V$ volume of the gas which contains $n$ number of molecules at temperature $T$and is given as $PV=nRT$
However, the ideal gas law doesn’t give any information of the nature of reaction, i.e. when the gas is expanding or compressing does it absorb heat or release heat. Also as the name suggests these gases are ideal and such gases don’t exist in the real world they are hypothetical in nature.
Here, given that the atmospheric pressure is $100\;kPa$, and the mass of gas is $8\;Kg$ and the piston area is $60cm^{2}$
Then the total pressure $P=P_{0}+\dfrac{m}{A}$
$\implies P=10^{5}+\dfrac{8\times 9.8}{60\times 10^{-4}}= 1.13\times 10^{5}N/m^{2}$
We also know from the first law of thermodynamics$\Delta Q=\Delta U+\Delta W$, where $Q$ is the heat, $U$ is the internal energy and $W$ is the work done by the system.
We also know that $\Delta W=P\Delta V$ and $\Delta U=nC\Delta T$
Here, given that when the gas is heated from $30^{\circ}C$ to $100^{\circ}C$ the piston rises $20\;cm$, then the $\Delta V=0.2\times 60\times 10^{-4}m^{3}=12\times 10^{-4}m^{3}$
$\implies \Delta Q_{1}=\Delta U_{1}+\Delta W_{1}$
$\implies \Delta Q_{1}=nC(100^{\circ}C-30^{\circ}C)+1.13\times 10^{5}\times 12\times 10^{-4}m^{3}$
$\implies \Delta Q_{1}=nC(100^{\circ}C-30^{\circ}C)+136J$
When there is no change in volume, and the gas is cooled back to$30^{\circ}C$ from $100^{\circ}C$, then $\Delta V=0$
$\implies \Delta Q_{2}=\Delta U_{2}+\Delta W_{2}$
$\implies Q_{2}=nC(30^{\circ}C-100^{\circ}C)+0J$
Also, given that $Q_{1}-Q_{2}=(130+x)$
$\implies Q_{1}-Q_{2}=nC(100^{\circ}C-30^{\circ}C)+136J-nC(30^{\circ}C-100^{\circ}C)+0J$
$\implies Q_{1}-Q_{2}= 136$
$\implies 136=(130+x)$
$\implies x=6J$

Therefore, the required value of x is 6J.

Note:
From ideal gas law, we know that $PV=nRT$ where $P$ is the pressure applied on the and $V$ is the volume of the gas which contains $n$ number of molecules at temperature $T$ and $R$ is the gas constant. We can vary the different parameters to understand the behaviours of the gas in various conditions.