Question

What is the final pressure (expressed in atm) of a 3.05 l system initially at 724 mm hg and 298 k, that is compressed to a final volume of 2.60 l at 273 k?

Answer 1

To answer this question, we will use the following equation:
(P1V1) / T1 = (P2V2) / T2 where:
P1 = 724 mm hg
V1 = 3.05 L
T1 = 298 K
P2 is the unknown we want to find
V2 = 2.6 L
T2 = 173 K

Substitute with the givens to calculate P2 as follows:
(724*3.05) / 298 = (P2*2.6) / 273
P2 = 778.057047 mm Hg = 1.023759 atm

Answer 2

$\boxed{1.0237596\;{\text{atm}}}$ is the final pressure of a 3.05 L system initially at 724 mm Hg ad 298 K that is compressed to a final volume of 2.60 L at 273 K.

Further Explanation:

An ideal gas is a hypothetical gas that is composed of 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 the equation of state for any hypothetical gas. The expression for the ideal gas equation is as follows:

$\boxed{{\mathbf{PV = nRT}}}$                                                   ...... (1)

Here,

P is the pressure of the gas.

V is the volume of the gas.

T is the absolute temperature of the gas.

n is the number of moles of gas.

R is the universal gas constant.

Rearranging equation (1), we get:

$\frac{{PV}}{T} = nR$                                                           ...... (2)

For a particular gas, the number of moles (n) and the universal as constant (R) both are constants.

If a specific gas with ${P_1}$, ${V_1}$ and ${T_1}$ is subjected to any change and the final parameters being ${P_2}$, ${V_2}$ and ${T_2}$. So equation (2) becomes,

$\frac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P_2}{V_2}}}{{{T_2}}}$               ...... (3)

Here,

${P_1}$ is the initial pressure of the gas.

${V_1}$ is the initial volume of the gas.

${T_1}$ is the initial temperature of the gas.

${P_2}$ is the final pressure of the gas.

${V_2}$ is the final volume of the gas.

${T_2}$ is the final temperature of the gas.

Calculation of the final pressure $\left( {{{\mathbf{P}}_{\mathbf{2}}}} \right)$ of the gas

Rearranging equation (3), we get:

${P_2} = \frac{{{P_1}{V_1}{T_2}}}{{{T_1}{V_2}}}$                             ...... (4)

We have ${P_1} = 724\;{\text{mm Hg}}$

${V_1} = 3.05\;{\text{L}}$

${T_1} = 298\;{\text{K}}$

${V_2} = {\text{2}}{\text{.60 L}}$

${T_2} = {\text{273 K}}$

Substitute these values in equation (4).

$\begin{gathered}{P_2}=\frac{{\left( {724\;{\text{mm Hg}}} \right)\left( {3.05\;{\text{L}}} \right)\left( {273\;{\text{K}}} \right)}}{{\left( {298\,{\text{K}}} \right)\left( {2.60\;{\text{L}}} \right)}} \\= 778.05705\;{\text{mm Hg}} \\ \end{gathered}$

Conversion Factor:

$1\;{\text{mm Hg}} = {\text{0}}{\text{.00131579 atm}}$

So the value of ${P_2}$ (in atm) is calculated as follows:

$\begin{gathered}{P_2}=\left( {778.05707\;{\text{mm Hg}}}\right)\left( {\frac{{{\text{0}}{\text{.00131579 atm}}}}{{1\;{\text{mm Hg}}}}} \right) \\= 1.023759\;{\text{atm}}\\\end{gathered}$

So the final pressure of the gas is 1.023759 atm.

Learn more:  

1. Rate of chemical reaction: https://brainly.com/question/1569924

2. Chemical bonds in NaCl: https://brainly.com/question/5008811

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 gas, initial, final, P, V, T, P1, P2, V1, V2, T1, T2.