Question

An ideal gas undergoes a cyclic process abcda which is shown by pressure-density curve.


Then,

• A. work done by the gas in process bc is zero
• B. work done by the gas in the process cd is negative
• C. internal energy of the gas at point 'a' is greater than the internal energy at point 'c'
• D. net work done by the gas in the cycle is negative

Answer 1

Answer: (A), (B), (D)

The correct options are
A work done by the gas in process bc is zero
B work done by the gas in the process cd is negative
D net work done by the gas in the cycle is negative

$P=\frac{\rho }{M_0}RT\Rightarrow \frac{P}{\rho }=\frac{R}{M_0}T\mathrm{Slope}\mathrm{of}\mathrm{the}\mathrm{curve}\mathrm{is}\mathrm{proportional}\mathrm{to}\mathrm{temperature}\mathrm{Hence},\mathrm{cd}\mathrm{and}\mathrm{ab}\mathrm{are}\mathrm{isothermal}\mathrm{processes}.\text{As}\rho \propto \frac{1}{V},\mathrm{bc}\mathrm{and}\mathrm{da}\mathrm{are}\mathrm{isochoric}\mathrm{processes}.$

• Since is no change of volume during process bc, the work done is zero.
• Area under the curve cd gives the work done, and volume is decreasing. So, work done is negative.
• Net work done is negative, from the PV diagram (because the arrows are anticlockwise)
• Internal energy is proportional to temperature. Since ab is an isothermal process, we can compare temperature at c and b. $T_a=T_b<T_c$ $⟹U_a<U_c$