Question

At NTP one mole of diatomic gas is compressed adiabatically to half of its volume $\gamma = 1.41$. The work done on gas will be
(a) $1280J$
(b) $1610J$
(c) $1818.67J$
(d) $2025J$

Answer 1

Hint: Use the relation between temperature and volume due to compression adiabatically to find the temperature after compression.
Use the formula of work done by the gas due to adiabatic compression in terms of temperature and calculate the amount of work done by the gas. Here, the value of the universal gas constant is needed.

Formula used: For the volumes \[{V_1}\] and ${V_2}$ at initial and final temperature ${T_1}$ and ${T_2}$ due to adiabatic compression $\dfrac{{{T_2}}}{{{T_1}}} = {\left( {\dfrac{{{V_1}}}{{{V_2}}}} \right)^{\gamma - 1}}$
The work done by the gas, $W = \dfrac{{R\left( {{T_1} - {T_2}} \right)}}{{\gamma - 1}}$
$R$ is the universal gas constant.

Complete step-by-step solution:
For a gas compressing under an adiabatic process, the relation between temperature and volume is \[T{V^{\gamma - 1}} = k\] , where \[k\] is some constant.
Using this relation,
For the volumes \[{V_1}\] and ${V_2}$ at initial and final temperature ${T_1}$ and ${T_2}$ due to adiabatic compression $\dfrac{{{T_2}}}{{{T_1}}} = {\left( {\dfrac{{{V_1}}}{{{V_2}}}} \right)^{\gamma - 1}}................(1)$
Given that, ${T_1} = 273K$ [since in N.T.P]
${V_2} = \dfrac{{{V_1}}}{2}$
$\gamma = 1.41$
So, from eq (1) ${T_2} = {T_1}{\left( {\dfrac{{{V_1}}}{{\dfrac{{{V_1}}}{2}}}} \right)^{\gamma - 1}}$
$ \Rightarrow {T_2} = 273 \times {2^{(1.41 - 1)}}$
$ \Rightarrow {T_2} = 363K............(2)$
Now, The work done by the gas due to the adiabatic process,
$W = \dfrac{{R\left( {{T_1} - {T_2}} \right)}}{{\gamma - 1}}.............(3)$
$R$ is the universal gas constant, $R = 8.31$
Putting the calculated and given values in the eq (3), we get
\[W = \dfrac{{8.31\left( {273 - 362.73} \right)}}{{1.41 - 1}}\]
\[ \Rightarrow W = - \dfrac{{8.31 \times 89.73}}{{0.41}}\]
\[ \Rightarrow W = - \dfrac{{745.6563}}{{0.41}}\]
\[ \Rightarrow W = - 1818.67\]
So, the work done \[ \Rightarrow W = - 1818.67J\]
Option (c) is the correct answer.

Note: An adiabatic process during which no heat is gained or lost by the system. The first law of thermodynamics with Q=0 shows that every one of the modifications in internal energy is within the type of work done. This puts a constraint on the warmth engine method resulting in the adiabatic condition. This condition is often accustomed to derive the expression for the work done throughout an adiabatic process.