Question

Work done by $0.1$ mol of a gas ${27^ \circ }C$ to double its volume at constant pressure is? $\left( {R = 2calmo{l^{ - 1}}{K^{ - 1}}} \right)$
A. $54cal$
B. $600cal$
C. $60cal$
D. $546cal$

Answer 1

Hint: At constant pressure the volume of an ideal gas is directly proportional to the absolute temperature, this is Charles law. By using this law we can find out the temperature when the volume of the gas becomes double. Thereby the work done can be calculated by taking the formula $Work done = W = P\Delta V$.

Complete answer:
Since, the volume of gas is doubled at constant pressure as given in the question, therefore, ${V_2} = 2{V_1}$(given)

Initial Temperature of gas,
${T_1} = {27^ \circ }C = 300K$(given)
$\left( {^ \circ C + 273 = K} \right)$

Now, we know that at constant pressure, the volume of gas is directly proportional to its absolute temperature (according to Charles Law).
i.e., $\dfrac{V}{T} = $constant
or, $\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{T_1}}}{{{T_2}}}$ … (1)

Substituting the required values in eq. (1), we get
$ \Rightarrow \dfrac{{{V_1}}}{{2{V_1}}} = \dfrac{{{T_1}}}{{{T_2}}}$
$ \Rightarrow {T_2} = 2{T_1}$

Then the change in temperature will be
$\Delta T = {T_2} - {T_1} = 2{T_1} - {T_1} = {T_1} = 300K$

Also, we know that Pressure-Volume Work in thermodynamics is defined as: -
At constant pressure, $Workdone = W = P\Delta V$ … (2)
But Ideal-Gas Equation is given as $P\Delta V = nR\Delta T$ … (3)

Equating (2) and (3), work done will be: -
$W = nR\Delta T$ … (4)
where $R = 2calmo{l^{ - 1}}{K^{ - 1}}$ = Gas Constant
and, $n = 0.1$ = no. of moles (given)

Substituting these values in eq. (4), we get
$W = 0.1 \times 2 \times \Delta T$

Substituting the value of $\Delta T$ in the above expression, we get
$ \Rightarrow W = 0.1 \times 2 \times 300$
$ \Rightarrow W = 60cal$
Thus, the work done by gas to double its volume is about $60cal$.

Hence, the correct option is (C) $60cal$.

Note: Since this is a problem based on the condition of constant pressure hence, charle’s law is used here to establish a relationship between the volume and the temperature. Always keep in mind to use these mathematical relationships to determine the solution i.e., while calculating the workdone by the gas.