Question

At a constant temperature a gas is initially at 2 atm pressure. To compress it to \[{\dfrac{1}{8}^{th}}\] of initial volume, pressure to be applied is:
A. 4 atm
B. 8 atm
C. 12 atm
D. 16 atm

Answer 1

Hint: We know the initial pressure of a gas given in the question as ${P_1}$ and let applied pressure be ${P_2}$ which we have to find out and the value of final volume to the initial volume is given by, $\dfrac{{{V_2}}}{{{V_1}}} = \dfrac{1}{8}$. Using the equation of Boyle’s Law, ${P_1}{V_1} =
{P_2}{V_2}$, we can easily calculate the value of applied pressure $({P_2})$ by substituting the
values in the equation.

Complete step by step answer:
Given in the question is,
Initial pressure of the gas = 2 atm
Final volume by initial volume, $\dfrac{{{V_2}}}{{{V_1}}} = \dfrac{1}{8}$
The final pressure of the gas =?
By using Boyle’s Law,
${P_1}{V_1} = {P_2}{V_2}$
Where,
${P_1} = $ Initial pressure
${P_2} = $ Final pressure
${V_1} = $ Initial volume
${V_2} = $ Final volume
Putting the value in the above equation, we get
$2 \times {V_1} = {P_2}{V_2}$
$\Rightarrow$ $2 = {P_2} \times \dfrac{{{V_2}}}{{{V_1}}}$
$\Rightarrow$ $2 = {P_2} \times \dfrac{1}{8}$
$\Rightarrow$ ${P_2} = 16atm$
Hence, the applied pressure ${P_2}$ is equal to 16 atm.

Therefore, the correct answer is option (D).

Note: Boyle’s law, which is also known as Mariotte’s law, is a relation that concerns the compression and expansion of a gas at a constant temperature. This empirical relation was formulated by the physicist Robert Boyle in 1662. According to Boyle’s Law, the pressure (p) of a given quantity of gas varies inversely with its volume (v) at constant temperature; that is, in equation form it is, \[pv\; =
\;k\], a constant. The relationship was also discovered by the French physicist Edmo Mariotte in 1676.