Question

A 325 ml sample of gas is initially at a pressure of 721 torr and a temperature of 32°c. if this gas is compressed to a volume of 286 ml and the pressure increases to 901 torr, what will be the new temperature of the gas (reported to three significant figures in °c)?

Answer 1

You can solve this question by using the ideal gas PV=nRT formula. Be careful though because the temperature used in the equation is kelvin, not celcius. The formula to determine the temperature would be: T= PV/nR

The calculation would be:
T1/T2= (P1V1/nR) / (P2V2/nR)
32+273.15/T2= (325*721) /(286*901)
305.15/T2= 0.90934
T2= 305.15/ 0.90934= 335.57 K= 62.4C

Answer 2

The new temperature of the given gas is $\boxed{62.4\:^{\circ}\text{C}}$.

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.

The expression for the ideal gas equation is as follows:

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

Here, P is the pressure of gas.

V is the volume of the gas.

T is the absolute temperature of gas.

n denotes the number of moles of gas.

R is the universal gas constant.

Rearranging equation (1), we get:

$\frac{{PV}}{T} = nR$

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 ${T_1}$  as initial parameters is treated to the final parameters being ${P_2}$ , ${V_2}$  and ${T_2}$ . So equation (1) becomes

$\frac{{{P_1}{V_1}}}{{{T_1}}} = \frac{{{P_2}{V_2}}}{{{T_2}}}$                …… (2)

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 temperature $\left({{{\mathbf{T}}_{\mathbf{2}}}}\right)$ of the gas:

Rearranging equation (2), we get:

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

Firstly, ${T_1}$ is to be converted into K. The conversion factor for this is,

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

So ${T_1}$  is calculated as follows:

$\begin{aligned}{\text{Temperature}}\left( {\text{K}} \right)&= \left({32 + 273}\right)\;{\text{K}}\\&=305\;{\text{K}}\\\end{aligned}$

The value of ${P_2}$  is 901 torr.

The value of ${V_2}$  is 286 mL.

The value of ${T_1}$  is 305 K.

The value of ${P_1}$  is s721 torr.

The value of ${V_1}$  is 325 mL.

Substitute these values in equation (3).

$\begin{aligned}{T_2}&=\frac{{\left({{\text{901 torr}}}\right)\left({{\text{286 mL}}}\right)\left({{\text{305 K}}}\right)}}{{\left({{\text{721 torr}}}\right)\left({{\text{325 mL}}}\right)}}\\&={\text{335}}{\text{.40693 K}}\\\end{aligned}$

Conversion Factor:

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

So the value of ${T_2}$ $\left({^\circ{\text{C}}}\right)$  is calculated as follows:

$\begin{aligned}{T_2}&=\left( {{\text{335}}{\text{.40693}}-{\text{273}}}\right){\text{ }}^\circ {\text{C}}\\&={\text{62}}{\text{.40693 }}^\circ{\text{C}}\\&\approx{\mathbf{62}}{\mathbf{.4 ^\circ C}}\\\end{aligned}$

Learn more:

1. Law of conservation of matter states: https://brainly.com/question/2190120

2. Calculation of volume of gas: https://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 gas, initial, final, K, conversion factor, 325 mL, 286 mL, 721 torr, 901 torr, 62.4 degree Celsius.