Question

A bulb with chlorine gas at ambient pressure contains $ 3.55 $ g at $ {t^ \circ } $ C. When the bulb is placed in a second thermostat at a temperature $ {30^ \circ } $ C higher and the stopcock of the bulb temporarily opened (and closed again) to restore the initial pressure, the bulb is now found to contain $ 3.2 $ g of chlorine. What is the value of $ {t^ \circ } $ C?
(A) $ {2.81^ \circ } $ C
(B) $ {8.21^ \circ } $ C
(C) $ {8.12^ \circ } $ C
(D) $ {1.28^ \circ } $ C

Answer 1

Hint: Ideal gas law is an equation of state of ideal gas. Ideal gas is a hypothetical gas. Ideal gas is a theoretical gas composed of many randomly moving point particles. To solve this question, we will use the ideal gas equation. We shall use the initial and final condition and equate them to calculate the temperature.

Formula used:
PV = nRT
where P is the pressure, V is the volume, T is the temperature, n is the moles of the gas or the amount of substance and R is the ideal gas constant.

Complete Step by step solution
The ideal gas law, also called as the general gas equation is given as follows,
PV = nRT
where P is the pressure, V is the volume, T is the temperature, n is the moles of the gas or the amount of substance and R is the ideal gas constant. The value of R is constant for all the gases.
The molecular formula for chlorine gas is $ C{l_2} $ . Hence the molecular mass of chlorine gas will be twice the atomic mass of one chlorine atom and that will be equal to 71.
Let $ {n_1} $ be the initial number of moles of chlorine gas and $ {n_2} $ be the final number of chlorine gas. Let $ {T_1} $ be the initial temperature and $ {T_2} $ be the final temperature. The number of moles is equal to the given weight divided by the molecular mass.
Hence,
 $ {T_1} = \left( {t + 273} \right)K $ and $ {T_2} = (t + 273 + 30)K = \left( {t + 303} \right)K $
 $ {n_1} = \dfrac{{3.55}}{{71}} $ and $ {n_2} = \dfrac{{3.2}}{{71}} $
In the given situation, pressure and volume remain constant. Hence,
 $ {n_1}{T_1} = {n_2}{T_2} $
Substituting the values,
 $ \dfrac{{3.55}}{{71}}\left( {t + 273} \right) = \dfrac{{3.2}}{{71}}\left( {t + 303} \right) $
Cancelling the same denominator on both sides and calculating we get,
 $ t\left( {3.55 - 3.2} \right) = \left( {3.2 \times 303} \right) - \left( {273 \times 3.55} \right) $
 $ \therefore t = \dfrac{{\left( {3.2 \times 303} \right) - \left( {273 \times 3.55} \right)}}{{\left( {3.55 - 3.2} \right)}} = {1.28^ \circ }C $
Therefore, the value of t is $ {1.28^ \circ } $ C. The correct answer is option D.

Note
The ideal gas equation relates Boyle’s law, Charles law and Avogadro’s law together.
Boyle’s law is the pressure of a given mass of an ideal gas is inversely proportional to its volume at a constant temperature. Charles law is the experimental law when pressure is constant, temperature and volume are in direct proportion.