Answer to Question #134380 in Mechanics | Relativity for Jessica

Question

An ideal gas at 10.3 °C and a pressure of 2.48 x 10^5 Pa occupies a volume of 3.12 m3. (a) How many moles of gas are present? (b) If the volume is raised to 5.45 m3 and the temperature raised to 25.6 °C, what will be the pressure of the gas?

Answer 1

The ideal gas equation is given by

$pV=nRT$ ….........(1)

where $p$ is the pressure in $Pa$

$V$ is the volume in $m^3$

n is the number of mole

$R = 8.314\ J.mol^{-1}.K^{-1}$ is the universal gas constant and

$T$ is the absolute temperature in $K$ of the gas.

(a) Given, $p=2.48\times 10^5\ Pa,\ V=3.12\ m^3, T=10.3 \degree C=(273+10.3)K=283.3K$

From Eq.(1), $n=\frac{pV}{RT}$

Substituting the given values of $p,V,R,T$, we get

$n=\frac{2.48\times 10^5\times 3.12}{8.314\times 283.3}\ mol$

$\therefore n=328.5\ mol$

(b) New volume $V'=5.45\ m^3$

New temperature $T'=25.6\degree C=(273+25.6)K=298.6K$

The new pressure is given by

$p'=\frac{nRT'}{V'}$

$p'=\frac{328.5\times 8.314\times 298.6}{5.45}\ Pa$

$\therefore p'=1.50\times 10^5\ Pa$

Answer: (a) There are 328.5 moles of gas.

(b) The new pressure of the gas will be $1.50\times 10^5\ Pa$ .