Question

A mass of diatomic gas $\gamma = 1.4$ at a pressure of 2 atmospheres is compressed adiabatically so that its temperature rises from ${27^\circ }C$ to ${927^\circ }C$. The pressure of the gas in the final state is
A. 28 atm
B. 68.7 atm
C. 256 atm
D. 8 atm

Answer 1

In order to solve this question we will use the gas equation for adiabatic process which is represented as $p{V^\gamma } = {\text{constant}}$. So first we will convert the given temperature from Celsius to Kelvin. And further substitute the values
Formula used- $p{V^\gamma } = {\text{constant,}}pV = RT$

Complete step-by-step solution:
Given that ${p_1} = 2atm$
Start temperature=${T_1} = {27^\circ }C$
End temperature=${T_2} = {927^\circ }C$
Let us convert the temperature in Kelvin by using the conversion formula
$T(k) = T(C) + 273$
Start temperature=${T_1} = 27 + 273 = 300{\text{ k}}$
End temperature=${T_2} = 927 + 273 = 1200{\text{ k}}$
As it is an adiabatic process, the gas equation will be
$p{V^\gamma } = {\text{constant}}$
As we know that
pV = RT \\\ \Rightarrow V = \dfrac{{RT}}{p} \\\
Since R is universal gas constant, it can be ignored
$\Rightarrow V = \dfrac{T}{p}$
Now, substituting this value of V in the gas equation $p{V^\gamma } = {\text{constant}}$, we will get
p{V^\gamma } = {\text{constant}} \\\ \Rightarrow p{\left( {\dfrac{T}{p}} \right)^\gamma } = {\text{ constant}} \\\ \Rightarrow p\dfrac{{{T^\gamma }}}{{{p^\gamma }}} = {\text{ constant}} \\\ \Rightarrow \dfrac{p}{{{p^\gamma }}}{T^\gamma } = {\text{ constant}} \\\ \Rightarrow {p^{1 - \gamma }}{T^\gamma } = {\text{ constant}} \\\
So if we consider two simultaneous conditions, then we have
p_1^{1 - \gamma }{T_1}^\gamma = p_2^{1 - \gamma }{T_2}^\gamma \\\ \Rightarrow \dfrac{{p_1^{1 - \gamma }}}{{p_2^{1 - \gamma }}} = \dfrac{{{T_2}^\gamma }}{{{T_1}^\gamma }} \\\ \Rightarrow {\left( {\dfrac{{{p_1}}}{{{p_2}}}} \right)^{1 - \gamma }} = {\left( {\dfrac{{{T_2}}}{{{T_1}}}} \right)^\gamma } \\\ \Rightarrow \left( {\dfrac{{{p_1}}}{{{p_2}}}} \right) = {\left( {\dfrac{{{T_2}}}{{{T_1}}}} \right)^{\dfrac{\gamma }{{1 - \gamma }}}} \\\ \Rightarrow {p_1} = {p_2}{\left( {\dfrac{{{T_2}}}{{{T_1}}}} \right)^{\dfrac{\gamma }{{1 - \gamma }}}} \\\
This can also be written as
$\Rightarrow {p_2} = {p_1}{\left( {\dfrac{{{T_1}}}{{{T_2}}}} \right)^{\dfrac{\gamma }{{1 - \gamma }}}}$
Substitute all the values, we get
$\Rightarrow {p_2} = 2{\left( {\dfrac{{300}}{{1200}}} \right)^{\dfrac{{1.4}}{{1 - 1.4}}}} \\\ \Rightarrow {p_2} = 2{\left( {\dfrac{{300}}{{1200}}} \right)^{\dfrac{{1.4}}{{ - 0.4}}}} \\\ \Rightarrow {p_2} = 2{\left( {\dfrac{{1200}}{{300}}} \right)^{\dfrac{{1.4}}{{0.4}}}} \\\ \Rightarrow {p_2} = 2{\left( 4 \right)^{\dfrac{7}{2}}} \\\ \Rightarrow {p_2} = 2 \times 128 \\\ \Rightarrow {p_2} = 256atm$
Hence, the pressure of the gas in the final state is 256 atm. So, the correct answer is option C.

Additional information: Classical thermodynamics considers 3 main styles of thermodynamic processes by the modification during a system, cycles during a system, and flow processes. outlined by modification during a system, a thermodynamical method may be a passage of a thermodynamic system from associate degree initial to a final state of natural philosophy equilibrium. Power cycles and warmth pump cycles are 2 main kinds of thermodynamic cycles. Power cycles are loops that rework any heat input into a mechanical work output, whereas loops of warmth pumps transfer heat from low to high temperatures mistreatment mechanical work as input.

Note: An adiabatic process occurs between a thermodynamic system and its surroundings without transferring heat or mass. The adiabatic process, unlike an isothermal process, only transfers energy to the surroundings as work. Ideal gas law is a well-defined approximation of the conduct of many gasses under various conditions. Ideal Gas Equation is a combination of empirical laws such as the law of Charles, Boyle’s law, the law of Gay-Lussac, and the law of Avogadro.