Question

Assertion: Work done in reversible isothermal compression is positive.
Reason: In the above process work is done on the gas.

Answer 1

Hint: The concept of reversible and irreversible compression of gas is to be used in this question. Work is the scalar product of pressure and volume, or $W=\int{Pdv+VdP}$.

Complete answer:
In order to answer our question, we need to learn about isothermal compression. Let us know about the types of thermodynamic processes:
Isothermal process: It is the process in which temperature is kept constant means the temperature of the initial and final state of the system along with the entire path of the process is the same.
Isobaric process: It is the process in which pressure is kept constant for the entire process.
Isochoric process: It is the process in which process volume is kept constant.
Reversible process: It is a process or change which is carried out infinitely slowly through a series of steps so that the system and the surroundings always remain almost in equilibrium state. The process is conducted in such a manner that any moment it could be reversed by an infinitesimal change. The opposing force at each step of the process is infinitesimally smaller than the driving force. This means that the system can be brought back to initial state without leaving any effect on the surroundings.
Irreversible process: Processes other than reversible processes are known as irreversible processes as irreversible processes". These processes do not meet the requirements of reversibility. So it can be defined as "When a process goes from initial to final state in a single step and can not be carried out in the reverse order, it is said to be irreversible.
Now, let us come to our question .Consider a reversible expansion of ideal gas by volume dV, then for an isothermal process, dE=0. Hence, dQ=-dW.The total work done during expansion of gas from initial to final volume is:
\[\int{dW=\int\limits_{{{V}_{1}}}^{{{V}_{2}}}{-PdV}}=\,-\int\limits_{{{V}_{1}}}^{{{V}_{2}}}{\dfrac{nRT}{V}dV}=\,-nRT\ln \dfrac{{{V}_{2}}}{{{V}_{1}}}\]
So,$W=\,-2.303RT{{\log }_{10}}\dfrac{{{V}_{2}}}{{{V}_{1}}}$
Since the temperature is constant, so ${{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}},\,or\,\dfrac{{{V}_{2}}}{{{V}_{1}}}=\dfrac{{{P}_{1}}}{{{P}_{2}}}$, hence,${{W}_{rev}}=\,-2.303nRT{{\log }_{10}}(\dfrac{{{P}_{1}}}{{{P}_{2}}})$
The work done by the process is positive as it gets compressed, also work is done on the gas.

So, the assertion and reason are correct.

NOTE: In case the same change is carried out in an irreversible manner, then the work done on the system can be given by the expression:${{W}_{irrev}}=\,-{{P}_{ext}}({{V}_{2}}-{{V}_{1}})$