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Question: A gas is collected in a graduated tube over mercury. The volume of the gas at \({25^\circ }C\) is \(...

A gas is collected in a graduated tube over mercury. The volume of the gas at 25C{25^\circ }C is 60mL60mL and the level of mercury in the tube is 110mm110mm above the mercury level. The barometer reads 760mm760mm. Volume of gas at STP is
(A) 33mL33mL
(B) 58mL58mL
(C) 39mL39mL
(D) 47mL47mL

Explanation

Solution

As we are very well aware of gas laws including Boyle's law and Charles’ law that combine to give a relationship between the three variables which are pressure, temperature and volume.

Complete Step by step answer:
We know that gas laws are made to study the behaviour of gases which is much easier than solids and liquids. Boyle's law states that at constant temperature the pressure of a fixed amount of gas varies inversely with the volume of the gas and Charles’ law states that volume occupied by a given mass of gas is directly proportional to the absolute temperature of gas at constant pressure. When combined together these two laws gave a new relationship between the three variables which are pressure, temperature and volume. This is called combined gas law. This relationship of variables can be shown as:
P1V1T1=P2V2T2\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}
In the question we are given with the initial volume V1=60  mL{V_1} = 60\;mL, temperature T1=25C{T_1} = {25^\circ }C and pressure P1=110  mm{P_1} = 110\;mm and the barometer reading shows P=760  mmP = 760\;mm, so the effective pressure will be =760110=650  mm = 760 - 110 = 650\;mm which we will consider the P1{P_1}.

Now, at STP the variables should be changed according to stand pressure and temperature conditions so after converting we get:
P1=650  torr{P_1} = 650\;torr, V1=60  mL{V_1} = 60\;mLand T1=25+273=298C{T_1} = 25 + 273 = {298^\circ }C
Similarly, P2=760  torr{P_2} = 760\;torr, T2=273C{T_2} = {273^\circ }Cand volume is to be calculated.

Applying the above formula we get:
P1V1T1=P2V2T2\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}
650×60298=760×V2273\Rightarrow \dfrac{{650 \times 60}}{{298}} = \dfrac{{760 \times {V_2}}}{{273}}
On solving for V2{V_2}: V2=47  mL{V_2} = 47\;mL.

Therefore the correct answer is (D).

Note: The volume of a given mass of gas increases with increase in temperature and decrease with decrease in temperature at a constant pressure. The absolute temperature is taken when the given temperature is added with 273C{273^\circ }C temperature, hence this temperature is also known as the kelvin temperature scale or thermodynamic scale of temperature. And the standard pressure we used is given as the 760  mmHg=760  torr760\;mmHg = 760\;torr.