Question

One liter of a gas weighs $4g$ at $300K$ and $1atm$ . If the pressure is reduced to $0.75atm$ then the temperature at which one liter of the same gas weighs $2g$ is:
A.$600K$
B.$900K$
C.$450K$
D.$800K$

Answer 1

Hint:Ideal gas law is defined as the equation of the state of hypothetical ideal gas. It is given by the relation that pressure is directly proportional to the temperature and number of moles and inversely proportional to the volume of gas.
Formula used:
$PV = nRT$
where, $P = $ Pressure of the gas
 $V = $ Volume of the gas
$n = $ number of moles
$R = $ universal gas constant
$T = $ Temperature.

Complete step by step answer:
If we talk about the ideal gas equation, we will get the relation as: $PV = nRT$
where, $P = $ Pressure of the gas, $V = $ Volume of the gas, $n = $ number of moles, $R = $ universal gas constant, $T = $ Temperature.
Since the gas is the same then $R$ will also be the same.
Combined gas law is the combination of three laws: Boyle’s law, Charles law, and Gay Lusaac’s law.
It is the ratio of the product of pressure and volume and the absolute temperature of gas is equal to constant.
Combining all the gas laws we get,
${P_1}{V_1} = {n_1}{T_1}$ ……1
Similarly,
${P_2}{V_2} = {n_2}{T_2}$ …2
Equating equation 1 and equation 2 we get,
$\dfrac{{{P_1}{V_1}}}{{{P_2}{V_2}}} = \dfrac{{{n_1}{T_1}}}{{{n_2}{T_2}}}$
But, ${n_1} = \dfrac{{{m_1}}}{{{M_1}}}$ , ${n_2} = \dfrac{{{m_2}}}{{{M_2}}}$ where, $m = $ mass, $M = $ Molar mass
Substituting the value of ${n_1}$ and ${n_2}$ we get,
$\dfrac{{{P_1}{V_1}}}{{{P_2}{V_2}}} = \dfrac{{\dfrac{{{m_1}}}{{{M_1}}}{T_1}}}{{\dfrac{{{m_2}}}{{{M_2}}}{T_2}}}$
Since the gas is same, so the molecular weight will be also same: ${M_1} = {M_2}$
Therefore, we get the equation as,
$\dfrac{{{P_1}{V_1}}}{{{P_2}{V_2}}} = \dfrac{{{m_1}{T_1}}}{{{m_2}{T_2}}}$
Where,
${P_1} = $ Initial pressure, ${P_2} = $ Final pressure, ${V_1} = $ Initial volume, ${V_2} = $ Final volume, ${T_1} = $ Initial temperature, ${T_2} = $ final temperature, ${m_1} = $ mass at pressure ${P_1}$ , ${m_2} = $ mass at pressure ${P_2}$ .
By using this equation we can calculate the temperature ${T_2}$ .
$\dfrac{{{P_1}{V_1}}}{{{P_2}{V_2}}} = \dfrac{{{m_1}{T_1}}}{{{m_2}{T_2}}}$
b)Given data:
${P_1} = 1atm$, ${P_2} = 0.75atm$
${V_1} = 1L$,${V_2} = 1L$
${m_1} = 4g$${m_2} = 2g$
${T_1} = 300K$${T_2} = ?$
Soln:
$\dfrac{{{P_1}{V_1}}}{{{P_2}{V_2}}} = \dfrac{{{m_1}{T_1}}}{{{m_2}{T_2}}}$
Substituting the values in this equation we get,
$ \Rightarrow \dfrac{{1 \times 1}}{{0.75 \times 1}} = \dfrac{{4 \times 300}}{{2 \times {T_2}}}$
$ \Rightarrow \dfrac{{1 \times 2}}{{0.75}} = \dfrac{{1200}}{{{T_2}}}$
So on doing the simplification,we have
$ \Rightarrow \dfrac{2}{{0.75 \times 1200}} = \dfrac{1}{{{T_2}}}$
$ \Rightarrow \dfrac{2}{{900}} = \dfrac{1}{{{T_2}}}$
And hence again on solving,we have
$ \Rightarrow {T_2} = \dfrac{{900}}{2}$
$ \Rightarrow {T_2} = 450K$
Therefore, the correct answer is option C)$450K$ .

Note:
The combined gas law is also used when we talk about the gases at ordinary temperatures and pressures. At high temperatures and pressures the combined gas law becomes less accurate. This law is mainly used in thermodynamics and fluid mechanics.