Solveeit Logo

Question

Question: A quantity of gas is collected in a graduated tube over mercury. The volume of the gas at \({20^ \ci...

A quantity of gas is collected in a graduated tube over mercury. The volume of the gas at 20C{20^ \circ }{\text{C}} is 50.0 mL50.0{\text{ mL}} and the level of mercury in the tube is 100 mm100{\text{ mm}} above the outside mercury level. The barometer reads 750 mm750{\text{ mm}}. Volume at STP is:
A. 39.8 mL39.8{\text{ mL}}
B. 40 mL40{\text{ mL}}
C. 42 mL42{\text{ mL}}
D. 60 mL60{\text{ mL}}

Explanation

Solution

We are given that the level of mercury in the tube is 100 mm100{\text{ mm}} above the outside mercury level and the barometer reading is 750 mm750{\text{ mm}}. To find the level of mercury in the tube, subtract the given level from the barometer reading.

Complete answer:
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=(750100) mm{P_1} = \left( {750 - 100} \right){\text{ mm}}
P1=650 mm\Rightarrow {P_1} = 650{\text{ mm}}
Thus, the volume of the gas is 650 mm650{\text{ mm}}.
The temperature given is 20C{20^ \circ }{\text{C}}. Thus,
T1=20C+273{T_1} = {20^ \circ }{\text{C}} + 273
T1=293 K\Rightarrow {T_1} = 293{\text{ K}}
And the volume of the gas is V1=50.0 mL{V_1} = 50.0{\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\Rightarrow {V_2} = \dfrac{{{P_1}{V_1}{T_2}}}{{{T_1}{P_2}}}
V2=650 mm×50.0 mL×273 K293 K×760 mm\Rightarrow {V_2} = \dfrac{{650{\text{ mm}} \times 50.0{\text{ mL}} \times 273{\text{ K}}}}{{293{\text{ K}} \times 760{\text{ mm}}}}
V2=39.8 mL\Rightarrow {V_2} = 39.8{\text{ mL}}
Thus, the volume of gas at STP is 39.8 mL39.8{\text{ mL}}.

**Thus, the correct option is (A) 39.8 mL39.8{\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.