A cylinder of volume 30 , L contains oxygen a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The initial pressure $P_1 = 15 \, atm = 15 	imes 1.013 	imes 10^5 \, Pa \approx 15.2 	imes 10^5 \, Pa$. The initial temperature $T_1 = 27 + 273

Vedclass · Accepted Answer

$0.14$

Vedclass · Answer

(A) The initial pressure $P_1 = 15 \, atm = 15 	imes 1.013 	imes 10^5 \, Pa \approx 15.2 	imes 10^5 \, Pa$. The initial temperature $T_1 = 27 + 273 = 300 \, K$. The volume $V = 30 \, L = 30 	imes 10^{-3} \, m^3$. Using the ideal gas law $PV = nRT = \frac{m}{M_w} RT$,the initial mass $m_1 = \frac{P_1 V M_w}{R T_1}$. $m_1 = \frac{(15.2 	imes 10^5) 	imes (30 	imes 10^{-3}) 	imes (32 	imes 10^{-3})}{8.31 	imes 300} \approx 0.586 \, kg$. After withdrawal,$P_2 = 11 \, atm = 11.14 	imes 10^5 \, Pa$,$T_2 = 17 + 273 = 290 \, K$. The final mass $m_2 = \frac{P_2 V M_w}{R T_2} = \frac{(11.14 	imes 10^5) 	imes (30 	imes 10^{-3}) 	imes (32 	imes 10^{-3})}{8.31 	imes 290} \approx 0.447 \, kg$. The mass of oxygen taken out is $\Delta m = m_1 - m_2 = 0.586 - 0.447 = 0.139 \, kg \approx 0.14 \, kg$.

Similar Questions

What is the universal gas constant? Write its value and unit.

The number of air molecules in a $(5 \,m \times 5 \,m \times 4 \,m)$ room at standard temperature and pressure is of the order of

$A$ gas of mass $m$ undergoes expansion at a constant pressure $P$,resulting in graph $D$. Which graph will be obtained if the same gas of mass $2m$ undergoes expansion at a pressure of $P/2$?

At the top of a mountain,a thermometer reads $7^{\circ}C$ and a barometer reads $70 \, cm$ of $Hg$. At the bottom of the mountain,these read $27^{\circ}C$ and $76 \, cm$ of $Hg$ respectively. The ratio of the density of air at the top to that at the bottom is

Vedclass Test Series

Exam Paper Generator

Online Exam Module