Question

During an adiabatic compression, $830J$ of work is done on $2$ moles of a diatomic ideal gas to reduce its volume by 50%. The change in its temperature is nearly: $\left( {{\text{R = 8}}{\text{.3J }}{{\text{K}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}} \right)$
(A) 20K
(B) 14K
(C) 40K
(D) 33K

Answer 1

We need to state the formula for work done in an adiabatic compression of an ideal gas and then we need to assign the values given in the question. The question can also be solved by relating work to the change in internal energy.
Formula Used: The formulae used in the solution are given here.
Work done is given by, $W = \dfrac{{nR\Delta T}}{{\gamma - 1}}$ where $\gamma$ is the ratio of specific heat (ordinary or molar) at constant pressure and at constant volume, $n$ is the number of moles, $R$ is universal gas constant and $\Delta T$ is the change in temperature.
Work done is also given by $W = - \Delta U$ where $\Delta U$ is the change in internal energy during the adiabatic expansion.

Complete Step by Step Solution: When an ideal gas is compressed adiabatically $\left( {Q = 0} \right)$, work is done on it and its temperature increases; in an adiabatic expansion, the gas does work and its temperature drops. Adiabatic compressions actually occur in the cylinders of a car, where the compressions of the gas-air mixture take place so quickly that there is no time for the mixture to exchange heat with its environment. Nevertheless, because work is done on the mixture during the compression, its temperature does rise significantly.
Work done is given by, $W = \dfrac{{nR\Delta T}}{{\gamma - 1}}$ where $\gamma$ is the ratio of specific heat (ordinary or molar) at constant pressure and at constant volume, $n$ is the number of moles, $R$ is universal gas constant and $\Delta T$ is the change in temperature.
It has been given that $830J$ of work is done on $2$ moles of a diatomic ideal gas to reduce its volume by 50%. The value of universal gas constant is ${\text{R = 8}}{\text{.3J }}{{\text{K}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$ and the ratio of specific heat $\gamma = 1.4$ for ideal gases.
Assigning the values, $W = 830J$, $n = 2$, ${\text{R = 8}}{\text{.3J }}{{\text{K}}^{{\text{ - 1}}}}{\text{mo}}{{\text{l}}^{{\text{ - 1}}}}$, $\gamma = 1.4$ and the value of change in temperature is thus found by,
$830 = \dfrac{{2 \times 8.3 \times \Delta T}}{{1.4 - 1}}$
Thus change in temperature, $\Delta T = 20K$.

Hence the correct answer is Option A.

Note: In thermodynamics, an adiabatic process is a type of thermodynamic process which occurs without transferring heat or mass between the system and its surroundings. Unlike an isothermal process, an adiabatic process transfers energy to the surroundings only as work.
In an adiabatic process, $W = - \Delta U$ where $\Delta U$ is the change in internal energy during the adiabatic expansion.
$\Delta U = n{C_V}\Delta T$ where ${C_V}$ is the specific heat at constant volume.