An open flask at 27^{circ}C contains air. To | vedclass

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

Question

(C) For an open flask,pressure $(P)$ and volume $(V)$ remain constant.According to the ideal gas law,$PV = nRT$,which implies $n \propto \frac{1}{T}$

Vedclass · Accepted Answer

$177$

Vedclass · Answer

(C) For an open flask,pressure $(P)$ and volume $(V)$ remain constant.According to the ideal gas law,$PV = nRT$,which implies $n \propto \frac{1}{T}$ at constant $P$ and $V$.Given: $T_1 = 27^{\circ}C = 300 \, K$.Let the initial number of moles be $n_1 = n$.After heating,one-third of the air is expelled,so the remaining moles are $n_2 = n - \frac{n}{3} = \frac{2n}{3}$.Using the relation $\frac{n_1}{n_2} = \frac{T_2}{T_1}$:$\frac{n}{2n/3} = \frac{T_2}{300}$$\frac{3}{2} = \frac{T_2}{300}$$T_2 = \frac{3}{2} 	imes 300 = 450 \, K$.Converting to Celsius: $T_2 = 450 - 273 = 177^{\circ}C$.

List-$I$	List-$II$
$(1)$ Boyle's Law	$(A)$ $p \propto T$
$(2)$ Charles's Law	$(B)$ $V \propto n$
$(3)$ Gay-Lussac's Law	$(C)$ $p \propto \frac{1}{V}$
$(4)$ Avogadro's Law	$(D)$ $pV = nRT$
$(5)$ Ideal Gas Equation	$(E)$ $V = k_2T$

An open flask at $27^{\circ}C$ contains air. To what temperature $(^{\circ}C)$ should it be heated so that one-third of the air is expelled from the flask?

Similar Questions

At $273 \ K$ temperature and $76 \ cm \ Hg$ pressure,the density of a gas is $1.964 \ g \ L^{-1}$. The gas is

Explain the law relating pressure and temperature.

For $0.09 \, \text{mol}$ of $CO_2$ at $300 \, K$,the volume is $112.0 \times 10^{-3} \, m^3$ at a pressure of $2.0 \times 10^4 \, Pa$. What will be the volume at a pressure of $4.0 \times 10^4 \, Pa$?

What is the density of $CO_2$ at $100\,^oC$ and $800\,mm\,Hg$ pressure in $g/L$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module