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Question: At \({27^ \circ }{\text{C}}\), \(195{\text{ ml}}\) of a gas are collected in a graduated tube over \...

At 27∘C{27^ \circ }{\text{C}}, 195 ml195{\text{ ml}} of a gas are collected in a graduated tube over Hg{\text{Hg}}. When barometer reads the reading equal to 745 mm745{\text{ mm}}, the level of inside the tube is 60 mm60{\text{ mm}} above the outside Hg{\text{Hg}} level. Calculate volume of gas at STP.
A) 160 ml160{\text{ ml}}
B) 130 ml130{\text{ ml}}
C) 180 ml180{\text{ ml}}
D) 210 ml{\text{210 ml}}

Explanation

Solution

We are given that the level of mercury in the tube is 60 mm60{\text{ mm}} above the outside mercury level and the barometer reading is 745 mm745{\text{ mm}}. To find the level of mercury in the tube, subtract the given level from the barometer reading.

Complete step-by-step solution:
We will first calculate the level of mercury in the tube. To find the level of mercury in the tube, subtract the given level from the barometer reading. The level in the tube is the measurement of pressure. Thus,
P1=(745−60) mm{P_1} = \left( {745 - 60} \right){\text{ mm}}
P1=685 mm{P_1} = 685{\text{ mm}}
Thus, the volume of the gas is 685 mm685{\text{ mm}}.
The temperature given is 27∘C{27^ \circ }{\text{C}}. Thus,
T1=27∘C+273{T_1} = {27^ \circ }{\text{C}} + 273
T1=300 K{T_1} = 300{\text{ K}}
And the volume of the gas is V1=195 ml{V_1} = 195{\text{ ml}}.
We have to calculate the volume of gas at STP i.e. standard temperature and pressure.
The standard temperature and pressure means the temperature is 298 K298{\text{ K}} and the pressure is 1 atm1{\text{ atm}}.
Thus, T2=273 K{T_2} = 273{\text{ K}} and P2=1 atm=760 mm{P_2} = 1{\text{ atm}} = 760{\text{ mm}}.
We know the ideal gas equation is,
PV=nRTPV = nRT
Where, PP is the pressure of the ideal gas,
VV is the volume of the ideal gas,
nn is the number of moles of ideal gas,
RR is the universal gas constant,
TT is the temperature of the gas.
Thus,
P1V1T1=P2V2T2\dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}}
V2=P1V1T2T1P2{V_2} = \dfrac{{{P_1}{V_1}{T_2}}}{{{T_1}{P_2}}}
V2=685 mm×195 ml×273 K300 K×760 mm{V_2} = \dfrac{{685{\text{ mm}} \times 195{\text{ ml}} \times 273{\text{ K}}}}{{{\text{300 K}} \times 760{\text{ mm}}}}
V2=159.93 ml=160 ml{V_2} = 159.93{\text{ ml}} = 160{\text{ ml}}
Thus, the volume of gas at STP is 160 ml160{\text{ ml}}.

Thus, the correct option is (A), 160 ml160{\text{ ml}}.

Note: Ideal gas is a hypothetical gas. The ideal gas molecules do not attract or repel each other. The ideal gas law states that the pressure, temperature and the volume of a gas are related to each other. The interaction between molecules of an ideal gas is elastic collision or elastic collision with the container walls.