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Question: A quantity of hydrogen gas occupies a volume of \(30.0\,{\text{mL}}\) at a certain temperature and p...

A quantity of hydrogen gas occupies a volume of 30.0 mL30.0\,{\text{mL}} at a certain temperature and pressure. What volume would half this mass of hydrogen occupy at triple the absolute temperature if the pressure were one ninth that of the original gas?
A. 270 mL{\text{270}}\,{\text{mL}}
B. 90 mL{\text{90}}\,{\text{mL}}
C. 405 mL{\text{405}}\,{\text{mL}}
D. 137 mL{\text{137}}\,{\text{mL}}

Explanation

Solution

The volume after changing the variable can be identified by using the ideal gas equation. The ideal gas equation relates the pressure, volume with temperature, and the number of moles of the gas.

Formula used: PV = nRT

Complete step by step answer:
The ideal gas equation is as follows:
PV = nRT
Where,
P is the pressure.
V is the volume.
n is the number of mole of gas.
R is the gas constant.
T is the temperature.
The initial temperature, pressure and volume of the gas is as follows:
PiVi = niRTi{P_i}{V_i}\, = \,{n_i}R{T_i}
On rearranging the ideal gas equation for volume,
Vi = niRTiPi{V_i}\, = \,\dfrac{{{n_i}R{T_i}}}{{{P_i}}}
At a certain temperature and pressure, a quantity of hydrogen gas occupies a volume of 30.0 mL30.0\,{\text{mL}} so,
Vi = niRTiPi=30.0 mL{V_i}\, = \,\dfrac{{{n_i}R{T_i}}}{{{P_i}}} = 30.0\,{\text{mL}}......(1)
The final mass is half of the initial mass so, nf =ni2{n_f}\, = \dfrac{{{n_i}}}{2}
The final temperature is the triple of the initial absolute temperature so, Tf =3 Ti{T_f}\, = 3\,{T_i} .
The final pressure is one ninth that of the original gas, so, Pf =Pi9{P_f}\, = \dfrac{{{P_i}}}{9}.
We can write the ideal gas equation for final volume as follows:
Vf = nfRTfPf{V_f}\, = \,\dfrac{{{n_f}R{T_f}}}{{{P_f}}}
Substitute the values of final temperature, pressure and number of moles.
Vf = ni2 ×3 Ti×R9Pi{V_f}\, = \,\dfrac{{{n_i}}}{2}\, \times 3\,{T_i} \times R\dfrac{9}{{{P_i}}}
Vf =   niRTipi3 ×92{V_f}\, = \,\,\,\dfrac{{{n_i}R{T_i}}}{{{p_i}}}\dfrac{{3\, \times 9}}{2}
Substitute value of niRTipi\dfrac{{{n_i}R{T_i}}}{{{p_i}}} from equation(1)(1) .
Vf =   30.0 mL×3 ×92{V_f}\, = \,\,\,30.0\,{\text{mL}} \times \dfrac{{3\, \times 9}}{2}
Vf =  405 mL{V_f}\, = \,\,405\,{\text{mL}}
So, the final volume of the hydrogen gas is 405 mL405\,{\text{mL}}.

Therefore option (C) 405 mL{\text{405}}\,{\text{mL}} is correct.

Note: Volume of the ideal gas is directly proportional to the temperature so, on increasing the temperature volume increases and inversely proportional to the pressure so, on decreasing pressure the volume increases.