The densities of air at 25,^o C and 100,^o C | vedclass

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

Question

(B) According to Charles's law,$V \propto T$ at constant pressure. Since density $d = \frac{m}{V}$,we have $d \propto \frac{1}{T}$,which implies $d_1

Vedclass · Accepted Answer

$-200$

Vedclass · Answer

(B) According to Charles's law,$V \propto T$ at constant pressure. Since density $d = \frac{m}{V}$,we have $d \propto \frac{1}{T}$,which implies $d_1 T_1 = d_2 T_2$. Let the absolute zero be $x\,^o C$. Then $T_1 = (25 - x) 	ext{ K}$ and $T_2 = (100 - x) 	ext{ K}$. Given $d_1 = 1.2\ gL^{-1}$ and $d_2 = 0.9\ gL^{-1}$. Substituting the values: $1.2 	imes (25 - x) = 0.9 	imes (100 - x)$. $1.2(25) - 1.2x = 0.9(100) - 0.9x$. $30 - 1.2x = 90 - 0.9x$. $30 - 90 = 1.2x - 0.9x$. $-60 = 0.3x$. $x = \frac{-60}{0.3} = -200\,^o C$.

The densities of air at $25\,^o C$ and $100\,^o C$ are $1.2\ gL^{-1}$ and $0.9\ gL^{-1}$ respectively. Assuming that the air obeys Charles's law,the value of absolute zero of temperature in degree Celsius is .............. $^o C$.

Similar Questions

Match the following gas laws with their respective equations.

Under what conditions does a gas behave ideally? Why?

$A$ and $B$ are ideal gases. At $T(K)$,$2 \ L$ of $A$ with a pressure of $1 \ bar$ is mixed with $4 \ L$ of $B$ with a pressure $p_B \ bar$ in a $100 \ L$ flask. The pressure exerted by the gaseous mixture is $0.1 \ bar$. What is the value of $p_B$ in $bar$?

At a given temperature,the density of gas $x$ is twice the density of gas $y$. If the molar mass of $y$ is three times the molar mass of $x$,find the ratio of their pressures $(P_x/P_y)$.

If $2$ moles of an ideal gas at $546 \ K$ has a volume of $44.8 \ L$,then what will be its pressure (in $atm$)? $(R=0.082 \ L \ atm \ K^{-1} \ mol^{-1})$

Vedclass Test Series

Exam Paper Generator

Online Exam Module