Question

An engine takes in \[5\] moles of air at \[20^\circ C\]and \[1\]atm, and compresses it adiabatically to \[\frac{1}{{10}}th\]of the original volume. Assuming air to be a diatomic ideal gas made up of rigid molecules, the change in its internal energy during this process comes out to be \[X\]kJ. The value of \[X\]to the nearest integer is ___________.

Answer 1

Hint:An ideal gas has equation $PV = nRT$. An adiabatic process is a thermodynamic condition in which no heat change is observed. Poisson equations are used In case of an adiabatic reversible process for ideal gas that shows relationship between initial and final state variables. \[\Delta U = n{C_v}\Delta T\] is used to find out the internal energy of the system where \[\Delta U\;\]is the change in internal energy of system under observation, \[{C_v}\] is heat capacity at constant volume of the gas, R is the universal gas constant, \[\Delta T\]is the change in temperature of the gas and n is the no. of moles of the gas.

Complete answer:
No. of moles, \[n = {\text{ }}5\]
Initial Temperature \[{T_1} = {\text{ }}20^\circ C = 20 + 273 = {\text{ }}293K\]
Initial Pressure \[{P_1} = {\text{ }}1{\text{ }}atm\]
Let Initial volume = V
Therefore, Final volume= $\frac{V}{{10}}$

Internal energy requires change in temperature hence finding final temperature using Poisson’s equation as follow:
$T{V^{{\mathbf{\gamma }} - 1}} = CONSTANT$
${T_1}{V_1}^{{\mathbf{\gamma }} - 1} = {T_2}{V_2}^{{\mathbf{\gamma }} - 1}$
\[\gamma\] is known as Poisson ratio. It is the ratio of heat capacity at constant pressure- heat capacity at constant volume
\[{\mathbf{\gamma }} = \frac{{{C_p}}}{{{C_v}}}\]

For a diatomic gas, there are 3 translational degree of freedom and 2 rotational degree of freedom.
Hence value of \[{C_v} = \left( {3 + 2} \right){\text{ }}\frac{R}{2} = \frac{5}{2}R.\]
As \[{C_p} - {C_v} = R\]
Hence, \[{C_p} = R + {C_v}\]
\[Cp = {\text{ }}\frac{7}{2}{\text{ }}R\]
Thus, value of \[\gamma =\dfrac { 7}{5}\]
Putting in passion equation:
$(293){V_1}^{\frac{7}{5} - 1} = {T_2}{V_2}^{\frac{7}{5} - 1}$
\[{V_2} = \frac{{{V_1}}}{{10}}\]
\[(293){V_1}^{\frac{2}{5}} = {T_2}{\frac{{{V_1}}}{{10}}^{\frac{2}{5}}}\]
${T_2} = (293){10^{\frac{2}{5}}}$
${T_2} = 735.98$K
\[\Delta U = n{C_v}\Delta T\]
\[\Delta U = 5{C_v}({T_2} - {T_1})\]
\[\Delta U = 5(\frac{5}{2}R)(735.98 - 293)\]

Value of R= 8.314
\[\Delta U = 46,036.69J\]
\[\Delta U = 46KJ\]

Note: It is an application of first law of thermodynamics. These equations make use of temperature in kelvin only and use of Celsius unit will lead to error. The value of γ (Poisson ratio) is dependent on value of Cv and it is a sum of various degree of freedom. At high temperature, vibrational degree of freedom is considered. Thus, value of \[\gamma = \frac{9}{7}\]including one vibrational degree of freedom.