At NTP,the mass of one litre of air is 1.293 | vedclass

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

Question

(A) At $NTP$ (Normal Temperature and Pressure),the standard conditions are:Pressure $P = 1.013 	imes 10^5 \; Pa$Temperature $T = 273.15 \; K$Volume $

Vedclass · Accepted Answer

$0.29$

Vedclass · Answer

(A) At $NTP$ (Normal Temperature and Pressure),the standard conditions are:Pressure $P = 1.013 	imes 10^5 \; Pa$Temperature $T = 273.15 \; K$Volume $V = 1 \; L = 10^{-3} \; m^3$Mass $m = 1.293 \; g = 1.293 	imes 10^{-3} \; kg$Using the ideal gas equation $PV = m R_{specific} T$,where $R_{specific}$ is the specific gas constant:$R_{specific} = \frac{PV}{mT}$$R_{specific} = \frac{(1.013 	imes 10^5 \; Pa) 	imes (10^{-3} \; m^3)}{(1.293 	imes 10^{-3} \; kg) 	imes (273.15 \; K)}$$R_{specific} \approx \frac{101.3}{0.3531} \approx 287 \; J/(kg \cdot K)$Since the question asks for the unit $J/(K \cdot g)$,we convert $kg$ to $g$:$R_{specific} = \frac{287 \; J}{1000 \; g \cdot K} = 0.287 \; J/(K \cdot g) \approx 0.29 \; J/(K \cdot g)$.

At $NTP$,the mass of one litre of air is $1.293 \; g$. The value of the specific gas constant will be ...... $J/(K \cdot g)$.

Similar Questions

$A$ partition divides a container having insulated walls into two compartments $I$ and $II$. The same gas fills the two compartments. Based on the values given in the figure,the ratio of the number of molecules in compartments $I$ and $II$ is

If the pressure of an ideal gas contained in a closed vessel is increased by $0.5\%,$ the increase in temperature is $2 \, K.$ The initial temperature of the gas is ...... $^\circ C$

An ideal gas has volume ${V_0}$ at ${27^{\circ}C}$. It is heated at constant pressure so that its volume becomes $2{V_0}$. The final temperature is

The perfect gas equation for $4 \ g$ of hydrogen gas is

State and explain Dalton's law of partial pressure.

Vedclass Test Series

Exam Paper Generator

Online Exam Module