Question

An ideal gas is compressed isothermally until its pressure is doubled and then allowed to expand adiabatically to regain its original volume $(\gamma=1.4$ and $2^{-1.4}=0.38)$. The ratio of the final to initial pressure is
A. $0.76:1$
B. $1:1$
C. $0.66:1$
D. $0.86:1$

Answer 1

Ideal gas are gases that follow the ideal gas equation, which is $PV=nRT$. When this gas is compressed isothermally it follows the equation $P_{1}V_{1}=P_{2}V_{2}$, and when expanded adiabatically it follows the equation $P_{1}V_{1}^{\gamma}=P_{2}V_{2}^{\gamma}$

Formula used:
For an isothermal process $P_{1}V_{1}=P_{2}V_{2}$.
For adiabatic process $P_{1}V_{1}^{\gamma}=P_{2}V_{2}^{\gamma}$

Complete step-by-step answer:
Ideal gas are gases that follow the ideal gas equation, which is given by $PV=nRT$.
A process is isothermal in which the temperature remains constant. $\Delta T=0$. It is given by $PV=constant$. Similarly, a process is adiabatic when there is no transfer of heat or mass between the system and the surroundings during the process, $\Delta Q=0$. It is given as $PV^{\gamma}=constant$.
If the initial pressure, volume and temperature of the ideal gas is $P$,$V$ and $T$.
We know for the isothermal process $P_{1}V_{1}=P_{2}V_{2}$.
Given that, $P_{2}=2P$.
Then, when the gas undergoes isothermal compression $V_{2}=\dfrac{P_{1}V_{1}}{P_{2}}=\dfrac{PV}{2P}=\dfrac{V}{2}$
We know for adiabatic process $P_{1}V_{1}^{\gamma}=P_{2}V_{2}^{\gamma}$
Here, $P_{2}V_{2}^{\gamma}= P_{3}V_{3}^{\gamma}$
Given , $V_{3}=V$ and $\gamma=1.4$
Then, $2P(\dfrac{V}{2})^{1.4}=P_{3}V^{1.4}$
$\dfrac{P_{3}}{P}=(\dfrac{1}{2})^{1.4}\times 2$
Given that $2^{-1.4}=0.38$
Then $\dfrac{P_{3}}{P}=2\times 0.38=0.76$
Thus the ratio of final to initial pressure is $\dfrac{P_{3}}{P}=\dfrac{0.76}{1}$
Hence the answer is A.$0.76:1$.

Note: Since the gas is ideal, it can undergo isothermal and adiabatic processes. Here, to find the ratio of the pressure before and after the process, we can use the relationship between pressure and volume as given in isothermal and adiabatic equations. Note that for adiabatic process $\Delta Q=0$, so $P_{1}V_{1}^{\gamma}=P_{2}V_{2}^{\gamma}$ and for isothermal process $\Delta T=0$,so $P_{1}V_{1}=P_{2}V_{2}$.