To begin with Snape thinks we should review an ideal gas problem. What is the pressure exerted by 0.703 mol O2 in a 11.0 L conta

Question

Bas_tet [7]

4 years ago

8

To begin with Snape thinks we should review an ideal gas problem. What is the pressure exerted by 0.703 mol O2 in a 11.0 L conta

iner at 314 K. Express your answer in atmospheres.

Chemistry

2 answers:

Soloha48 [4] · Answer 1 · 2022-01-18T21:47:53+03:00

This can be solved using ideal gas equation:

ideal gas equation,

P\times V=n\times R\times T

Here,

P denotes pressure

V denotes volume

n denotes number of moles of gas

R denotes gas constant

T denotes temperature

The values given here are as follows;

T=314 K

R=0.0821 atm L mol ⁻¹

V=11.0 L

Number of moles of gas, n= 0.703 mol

Putting all the values in the above equation,

P=\frac{ 0.703\times 0.0821\times 314}{11.0 }

P=1.64 atm

So the pressure will be 1.64 atm.

Salsk061 [2.6K] · Answer 2 · 2022-01-18T23:27:42+03:00

The pressure exerted by 0.703 mol ${{\text{O}}_2}$ in a 11.0 L container at 314 K is $\boxed{{\text{1}}{\text{.647 atm}}}$ .

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 gas is as follows:

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

Here,

P is the pressure of the gas.

V is the volume of gas.

n denotes the number of moles of gas.

R is the gas constant.

T is the temperature of gas.

Rearrange the equation (1) to calculate P.

${\text{P}}=\dfrac{{{\text{nRT}}}}{{\text{V}}}$ ...... (2)

Given,

Volume of gas (V) is 11L.

Temperature of gas(T) is 314K.

Numbers of moles(n) are 0.703mol.

Universal gas constant R is $0.0821{\text{ atm}}\cdot{\text{L/mol}}\cdot{\text{K}}$ .

Substitute the values of V, T, R, n in equation (2) to calculate the pressure.

$\begin{aligned}{\text{P}}&=\dfrac{{\left( {{\text{0}}{\text{.703 mol}}} \right)\times\left( {0.0821{\text{ atm}} \cdot {\text{L/mol}}\cdot {\text{K}}}\right)\times\left( {{\text{314 K}}}\right)}}{{{\text{11 L}}}}\\&= {\text{1}}{\text{.647 atm}}\\\end{aligned}$