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Question: In a process, the pressure of the gas is inversely proportional to the square of the volume. If the ...

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

Explanation

Solution

The pressure of the gas is inversely proportional to the square of the volume i.e.
Given,p1V2Given,p \propto \dfrac{1}{{{V^2}}}
p=KV2\therefore p = \dfrac{K}{{{V^2}}}

Formula used: pV=nRTpV = nRT
Where nn= number of gas molecules,
RR= Universal gas constant,
TT= temperature of the gas in the kelvin unit.
Work done by the gas,
W=p.dVW = \int {p.dV}
dW=p.dV\Rightarrow dW = p.dV

Complete step by step answer: From Boyle's law we get p1Vp \propto \dfrac{1}{V}where pp=pressure of the gas and VV=volume of the gas.
And, from Charl’s law, we get VTV \propto Twhere TT=temperature of the gas in the kelvin unit.
Combining the two laws we get
pVTpV \propto T
pV=KT\Rightarrow pV = KT
This equation leads to the Ideal gas equation pV=nRTpV = nRT
Where nn= number of gas molecules,
RR= Universal gas constant.
According to the problem, The pressure of the gas is inversely proportional to the square of the volume i.e.
Given,p1V2Given,p \propto \dfrac{1}{{{V^2}}}
p=KV2............(1)\therefore p = \dfrac{K}{{{V^2}}}............(1)
Where PP=pressure of the gas in this process,
VV=volume of the gas,
KK= proportionality constant.
The ideal gas equation is,
pV=nRT.............(2)pV = nRT.............(2)
Put the value of pp from eq(1) in the eq(2) we get,
KV2×V=nRT\dfrac{K}{{{V^2}}} \times V = nRT
1V=nRTK\Rightarrow \dfrac{1}{V} = \dfrac{{nRT}}{K}
V1T\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=p.dVW = \int {p.dV}
Integrating the given equation
dW=p.dV\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.