Question

In a process, the pressure of the gas is inversely proportional to the square of the volume. If the temperature of the gas increase then the work done by the gas:
A) Is positive
B) Is negative
C) Is zero
D) May be positive

Answer 1

Hint:The pressure of the gas is inversely proportional to the square of the volume i.e.
\[Given,p \propto \dfrac{1}{{{V^2}}}\]
\[\therefore p = \dfrac{K}{{{V^2}}}\]

Formula used: \[pV = nRT\]
Where $n$= number of gas molecules,
$R$= Universal gas constant,
$T$= temperature of the gas in the kelvin unit.
Work done by the gas,
\[W = \int {p.dV} \]
\[ \Rightarrow dW = p.dV\]

Complete step by step answer:From Boyle's law we get \[p \propto \dfrac{1}{V}\]where $p$=pressure of the gas and $V$=volume of the gas.
And, from Charl’s law, we get \[V \propto T\]where $T$=temperature of the gas in the kelvin unit.
Combining the two laws we get
\[pV \propto T\]
\[ \Rightarrow pV = KT\]
This equation leads to the Ideal gas equation \[pV = nRT\]
Where $n$= number of gas molecules,
$R$= Universal gas constant.
According to the problem, The pressure of the gas is inversely proportional to the square of the volume i.e.
\[Given,p \propto \dfrac{1}{{{V^2}}}\]
\[\therefore p = \dfrac{K}{{{V^2}}}............(1)\]
Where $P$=pressure of the gas in this process,
$V$=volume of the gas,
$K$= proportionality constant.
The ideal gas equation is,
\[pV = nRT.............(2)\]
Put the value of $p$ from eq(1) in the eq(2) we get,
\[\dfrac{K}{{{V^2}}} \times V = nRT\]
\[ \Rightarrow \dfrac{1}{V} = \dfrac{{nRT}}{K}\]
\[ \Rightarrow V \propto \dfrac{1}{T}\]
Therefore, the volume of the gas will decrease if the temperature increases.
Now, Work done by the gas,
\[W = \int {p.dV} \]
Integrating the given equation
\[ \Rightarrow dW = p.dV\]
This means the change in work done is directly proportional to the change in volume of the gas for constant pressure.
Therefore, if the temperature increases the work done will be negative due to the decrease in volume of the gas.

Hence, option (B) is the correct one.

Notes:
Positive work is done on the gas when the gas is compressed that is work done by an external force on a system.
Negative work done by gas means expanding against external pressure that is work done by the system.