Question

Please solve this question​

Answer 1

Step-by-step explanation:

${\large{\mathbf{\underline{\red{Given :→}}}}}$

${\bold{\sf{➮ \: \: \: \: \: \: Volume \: of \: gas \: (V_1) \: = \: 30 \: liters \: = \: 30×{10}^{ - 3} m³}}}$

${\bold{\sf{➮ \: \: \: \: \: Pressure (P_1) \: = \: 15 atm \: = \:15 × 1.013 × 10⁵ \: Pa}}}$

${\bold{\sf{➮ \: \: \: \: \: Temperature (T_1) \: = \: 27°C \: = \: 300 K}}}$

${\bold{\sf{➮ \: \: \: \: \: Universal \: Gas \: Constant (R) \: = \: 8.314 J \: {mol}^{-1} {K}^{-1}}}}$

$\\ {\large{\mathbf{\underline{\red{To \: find :→}}}}}$

Mass of oxygen taken out of cylinder

$\\ {\large{\mathbf{\underline{\red{Solution :→}}}}}$

Let the initial number of moles of oxygen gas in the cylinder be n,

The gas equation is given as follows :

${\bold{\sf{P_1V_1 \: = \: n_1RT_1}}}$

Hence,

${\bold{\sf{n_1 \: = \: \frac{P_1V_1}{RT_1}}}}$

${\bold{\sf{: \: ⟹ \: \: \: \: \: \frac{15.195 \: × \: 10⁵ \: × \: 30 \: × \: {10}^{-3}}{8.314 \: × \: 300}}}}$

${\bold{\sf{: \: ⟹ \: \: \: \: \: 18.276}}}$

${\bold{\sf{ \: \: \: \: \: \: But \: n_1 \: = \: \frac{m_1}{M}}}}$

Where,

${\bold{\sf{: \: ⟹ \: \: \: \: \: m_1 \: = \: initial \: mass \: of \: oxygen}}}$

${\bold{\sf{: \: ⟹ \: \: \: \: \: M \: = \: molecular \: mass \: of \: oxygen \: = \: 32g}}}$

Thus,

${\bold{\sf{: \: ⟹ \: \: \: \: \: m_1 \: = \: N_1M}}}$

${\bold{\sf{: \: ⟹ \: \: \: \: \: 18.276 \: × \: 32}}}$

${\bold{\sf{: \: ⟹ \: \: \: \: \: 584.84 \: g}}}$

After sometime oxygen is withdrawn from the cylinder, the pressure and temperature reduced

${\bold{\sf{➮ \: \: \: \: \: Volume (V_2) \: = \: 30 \: liters \: = \: 30 \: × \: {10}^{-3}{m}^{3}}}}$

${\bold{\sf{➮ \: \: \: \: \: Pressure (P_2) \: = \: 11 atm \: = \: 11 \: × \: 1.013 \: × \: 10⁵ Pa }}}$

${\bold{\sf{➮ \: \: \: \: \: Temperature (T_2) \: = \: 17°C \: = \: 290 K}}}$

$\\ {\bold{\sf{Let \: n_2 \: be \: the \: number \: of \: moles \: left \: in \: the \: cylinder}}}$

The gas equation is given as :

${\bold{\sf{: \: ⟹ \: \: \: \: \: P_2V_2 \: = \: n_2RT_2}}}$

Hence,

${\bold{\sf{: \: ⟹ \: \: \: \: \: n_2 \: = \: \frac{P_2V_2}{RT_2}}}}$

${\bold{\sf{: \: ⟹ \: \: \: \: \: \frac{11.143 \: × \: 10⁵ \: × \: 30 \: × \: {10}^{-30}}{8.314 \: × \: 290}}}}$

${\bold{\sf{: \: ⟹ \: \: \: \: \: 13.86}}}$

But,

${\bold{\sf{: \: ⟹ \: \: \: \: \: n_2 = \: \frac{m_2}{M}}}}$

Where,

${\bold{\sf{m_2 \: is \: the \: mass \: of \: oxygen \: remain \: in \: the \: cylinder}}}$

Therefore,

${\bold{\sf{: \: ⟹ \: \: \: \: \: m_2 \: = \: n_2 × M \: = \: 13.86 × 32}}}$

${\bold{\sf{: \: ⟹ \: \: \: \: \: 453.1 \: g}}}$

Mass of oxygen taken out of the solution is given by the relation :

Initial mass of oxygen - final mass of oxygen

${\bold{\sf{: \: ⟹ \: \: \: \: \: m_1 \: - \: m_2}}}$

${\bold{\sf{: \: ⟹ \: \: \: \: \: 584.84g \: - \: 453.1g}}}$

We get,

${\bold{\sf{: \: ⟹ \: \: \: \: \: 131.74g}}}$

${\bold{\sf{: \: ⟹ \: \: \: \: \: 0.131kg}}}$

Hence 0.131 kg of oxygen is taken out of cylinder