Question

what is the volume of 2.8g co at 27° and 0.0821 atm pressure?​

Answer 1

Answer:

Explanation:

v=0.1*0.0821*300/0.0821=30

Answer 2

Answer:

Explanation:

Given,

Molar mass of CO(Carbon monoxide)$M==28 \text{ g mol}^{-1}$

Temperature in Kelvin scale = $T=27^\circ\text{ C}=(27+273)=300\text{ K}$

Atmospheric Pressure $P=0.0821\text{ atm}$

Weight $w=2.8\text{ g}$

Applying the ideal gas law we get

$PV=nRT\\\\\Rightarrow PV=\frac{\text{w}}{M}RT\\\\\Rightarrow V=\frac{\text{w}RT}{\text{MP}}\text{.......................equation 1}$

Where,

P is the pressure of the gas,

V is the volume of the gas,

R is the ideal, or universal, gas constant,

$R= 0.0821 \text{ L atm mol}^{-1} K^{-1}$

T is the absolute temperature of the gas.

w is the weight of the gas.

M is the Molar mass of the gas.

Therefore putting the value of all in equation 1 we get,

$V=\frac{\text{w}RT}{\text{MP}}\\\\V=\frac{2.8\times0.0821\times300}{28\times0.0821}=30$

Therefore Volume of the CO is 30 L