Question

What is the volume of 10.0 g of argon gas at 157°c and 2.50 kpa pressure?

Answer 1

Data Given:
                  Pressure = P =  2.5 kPa  =  0.02467 atm

                  Temperature = T = 157 °C + 273.15  =  430.15 K

                  Volume = V = ?

                  Mass = m = 10 g

First find out moles for given amount of Argon gas,
As,
                             Moles  =  Mass / M.mass
Putting values,
                             Moles  =  10 g / 39.94 g/mol  =  0.250 moles

Now using Ideal Gas Equation,

                                   P V  =  n R T
Solving for V,
                                   V  =  n R T / P
Putting Values,

         V  =  (0.25 mol × 0.0821 L.atm.mol⁻¹.K⁻¹ × 430.15 K) ÷ 0.02467 atm

                                          Volume =  357.87 L

Answer 2

$\boxed{{\text{357}}{\text{.75 L}}}$ of 10 g of argon gas is present at $157\;^\circ{\text{C}}$ and 2.50 kPa.

Further Explanation:

An ideal gas contains a large number of randomly moving particles that are supposed to have perfectly elastic collisions among themselves. It is just a theoretical concept, and practically no such gas exists. But gases tend to behave almost ideally at a higher temperature and lower pressure.

Ideal gas law is considered as the equation of state for any hypothetical gas. The expression for the ideal gas equation of argon is as follows:

${\text{PV}}={\text{nRT}}$                                 …… (1)

Here, P is the pressure of argon gas.

V is the volume of argon gas.

T is the absolute temperature of argon gas.

n is the number of moles of argon gas.

R is the universal gas constant.

Rearrange equation (1) to calculate the pressure of argon gas.

${\text{V}}=\frac{{{\text{nRT}}}}{{\text{P}}}$                                       …… (2)

Firstly, the temperature is to be converted into K. The conversion factor for this is,

${\text{0 }}^\circ{\text{C}}={\text{273 K}}$

So the temperature of argon is calculated as follows:

$\begin{gathered}{\text{Temperature}}\left({\text{K}}\right)=\left({157+273}\right)\;{\text{K}}\\=430\;{\text{K}}\\\end{gathered}$

The pressure is also to be converted into atm. The conversion factor for this is,

${\text{1 kPa}}={\text{0}}{\text{.00986923 atm}}$

So the pressure of argon gas is calculated as follows:

$\begin{gathered}{\text{Pressure of argon}}=\left({{\text{2}}{\text{.5 kPa}}}\right)\left({\frac{{{\text{0}}{\text{.00986923 atm}}}}{{{\text{1}}\;{\text{kPa}}}}}\right)\\={\text{0}}{\text{.02467 atm}}\\\end{gathered}$

The formula to calculate the moles of argon gas is as follows.

${\text{Moles of argon}}=\frac{{{\text{Given mass of argon}}}}{{{\text{Molar mass of argon}}}}$                        …… (3)

The given mass of argon gas is 10 g.

The molar mass of argon gas is 39.95 g/mol.

Substitute these values in equation (3)

$\begin{gathered}{\text{Moles of argon}}=\left({10\;{\text{g}}}\right)\left({\frac{{1\;{\text{mol}}}}{{39.95\;{\text{g}}}}}\right)\\={\text{0}}{\text{.250 mol}}\\\end{gathered}$

The pressure of argon gas is 0.02467 atm.

The volume of argon gas is 2 L.

The number of moles of argon gas is 0.250 mol.

The temperature of argon gas is 430 K.

Universal gas constant is 0.0821 L atm/K mol.

Substitute these values in equation (2).

$\begin{gathered}{\text{V}}=\frac{{\left({{\text{0}}{\text{.250 mol}}}\right)\left({{\text{0}}{\text{.0821}}\;{\text{L}}\cdot{\text{atm/mol}}\cdot{\text{K}}}\right)\left({{\text{430 K}}}\right)}}{{\left({{\text{0}}{\text{.02467 atm}}}\right)}}\\={\text{357}}{\text{.7523 L}}\\\approx{\mathbf{357}}{\mathbf{.75 L}}\\\end{gathered}$

Learn more:

1. Which statement is true for Boyle’s law: brainly.com/question/1158880

2. Calculation of volume of gas: brainly.com/question/3636135

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Ideal gas equation

Keywords: ideal gas, pressure, volume, absolute temperature, equation of state, hypothetical, universal gas constant, moles of argon gas, 0.250 mol,430 K, 273 K, P, V, n, R, T.