A home owner uses 4.00 times 10^{3} m^{3} of | vedclass

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

Question

(C) Given:Volume $V = 4.00 	imes 10^{3} \ m^{3} = 4.00 	imes 10^{3} 	imes 10^{3} \ L = 4.00 	imes 10^{6} \ L$Pressure $P = 1.0 \ atm$Temperature $

Vedclass · Accepted Answer

$26$

Vedclass · Answer

(C) Given:Volume $V = 4.00 	imes 10^{3} \ m^{3} = 4.00 	imes 10^{3} 	imes 10^{3} \ L = 4.00 	imes 10^{6} \ L$Pressure $P = 1.0 \ atm$Temperature $T = 300 \ K$Gas constant $R = 0.083 \ L \ atm \ K^{-1} \ mol^{-1}$Using the ideal gas equation $PV = nRT$,where $n = \frac{mass (m)}{Molar \ mass (M)}$:$n = \frac{PV}{RT} = \frac{1.0 	imes 4.00 	imes 10^{6}}{0.083 	imes 300} \ mol$$n = \frac{4.00 	imes 10^{6}}{24.9} \ mol \approx 1.6064 	imes 10^{5} \ mol$Molar mass of $CH_{4} = 12 + 4 	imes 1 = 16 \ g \ mol^{-1}$Mass $m = n 	imes M = 1.6064 	imes 10^{5} 	imes 16 \ g$$m = 25.7024 	imes 10^{5} \ g$Comparing with $X 	imes 10^{5} \ g$,we get $X = 25.7024$.Rounding to the nearest integer,$X = 26$.

Similar Questions

What is the value of $R$ in the ideal gas equation $pV = nRT$ in units of $bar \ L \ mol^{-1} \ K^{-1}$? Calculate it.

An ideal gas is heated in a rigid container so its temperature rises. Which is incorrect for this process?

An ideal gas will have maximum density when

$A$ hot air balloon has a volume of $2000 \text{ dm}^3$ at $99^\circ\text{C}$. What is the new volume if the air in the balloon cools to $80^\circ\text{C}$ (in $\text{ dm}^3$)?

The density of an ideal gas can be given by where $P$,$V$,$M$,$T$ and $R$ respectively denote pressure,volume,molar mass,temperature and universal gas constant.

Vedclass Test Series

Exam Paper Generator

Online Exam Module