Question

10 moles of ideal gas expand isothermally and reversibly from a pressure of 5 atm to 1 atm at 300 K. What is the largest mass which can be lifted through a height of 1 meter by this expansion?
A.4000 kg
B.4100 kg
C.4200 kg
D.4300 kg

Answer 1

Hint: Work done in an isothermal reversible process is given by the equation, \[W = - 2.303nRT\,In\left( {\dfrac{{{p_1}}}{{{p_2}}}} \right)\]. Since the number of moles, \[{P_{1,}}{P_2}\], Temperature is given, we can calculate the value of work done by substituting the value in the equation. After finding the work done, we can calculate the mass which can be lifted through a height of 1 m from the equation, \[W = Mgh\] and \[\left( {m = 9.8} \right)\].

Complete step by step answer:
Work done in this case is given by,
 $W = - 2.303nRT\,In\left( {\dfrac{{{p_1}}}{{{p_2}}}} \right)$
Given in the question,
Number of moles, \[n = 10\]
 \[R = 8.314J{K^{ - 1}}mo{l^{ - 1}}\]
 \[T = 300K\]
Initial pressure \[{p_1} = 5\] atm
Final pressure \[{p_2} = 1\] atm
We will now substitute the given values in the equation:
 $W = - 2.303nRT\,In\left( {\dfrac{{{p_1}}}{{{p_2}}}} \right)$
$\Rightarrow$ $W = - 2.303 \times 10 \times 8.314 \times 300 \times ln\dfrac{5}{1}$
$\Rightarrow$ \[W = - 2.303 \times 10 \times 8.314 \times 300 \times 0.698\] (Since, ln $\dfrac{5}{1} = 0.698$ )
$\Rightarrow$ W = 40094.115 J
Let Me be the mass which can be lifted through a height of 1 m
Work done in lifting the mass \[ = M \times g \times h\]
Height = 1 meter (given in the question)
 \[g = 9.8\,m{s^{ - 2}}\]
 \[W = Mgh\]

$\Rightarrow$ \[20047.057 = M \times 9.8 \times 1\]

$\Rightarrow$ $M = \dfrac{{40094 \times 115}}{{9.8 \times 1}} = 4091.236\,kg$
So, the largest mass which can be lifted through a height of 1 meter by this expansion is 4091.236 kg which is close to 4100 kg.

Therefore, the correct answer is option (B).

Note: We can write the unit of mass in Kg or g which is lifted through a height of 1 metre. Thus, we can calculate it in any of the two units. We should write the equation of the work done in terms of pressure, and thus we can determine the mass through it. We will not consider the negative sign in the equation as we know that the mass cannot be negative.