Question

A tank is filled with an ideal gas at 400 and pressure of 1.00 .

The tank is heated until the pressure of thegas in the tank doubles. What is the temperature of the gas?

	200 $\rm K$ 400 $\rm K$ 600 $\rm K$ 800 $\rm K$
Correct

Having been heated to 800 , at some point the tank starts to leak. By the time theleak is repaired, the tank is only half full, and the pressure ofthe remaining gas is again 1.00 . What is the temperature of the gas?

200 $\rm K$ 400 $\rm K$ 600 $\rm K$ 800 $\rm K$

Answer 1

Concepts and reason

The concept used to solve this problem is ideal gas law.

Initially, use the initial temperature, initial pressure, and final pressure to calculate the final temperature.

Finally, use the ideal gas law equation to calculate the temperature of the gas.

Fundamentals

Expression for the ideal gas law is,

$PV = nRT$

Here, P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

When number of moles and volume is constant the expression is,

$\frac{{{P_1}}}{{{T_1}}} = \frac{{{P_2}}}{{{T_2}}}$

Here, ${P_1}$ is the initial pressure, ${P_2}$ is the final pressure, ${T_1}$ is the initial temperature, and ${T_2}$is the final temperature.

Rearrange the above equation to get the final temperature of the gas,

${T_2} = \frac{{{P_2}{T_1}}}{{{P_1}}}$

Expression for the number of moles in the container before leakage of gas is,

$n = {n_1}$

Here, n is the number of moles and ${n_2}$ is the number of moles in the container before leakage of gas.

Expression for the number of moles in the container after leakage of gas is,

$n = \frac{{{n_1}}}{2}$

Expression for the ideal gas law before leakage of gas is,

${P_2}V = nR{T_2}$

Substitute ${n_1}$ for n.

${P_2}V = {n_1}R{T_2}$ …… (1)

Expression for the ideal gas law after leakage of gas is,

${P_3}V = nR{T_3}$

Here, ${P_3}$ is the pressure of the gas after leakage of gas and ${T_3}$ is the temperature of the gas after leakage of gas.

Substitute \[{{{n_1}} \mathord{\left/\\ {\vphantom {{{n_1}} 2}} \right.\\ \kern-\nulldelimiterspace} 2}\] for n.

${P_3}V = \left( {\frac{{{n_1}}}{2}} \right)R{T_3}$ …… (2)

(1)

The expression for the final temperature of the gas is,

${T_2} = \frac{{{P_2}{T_1}}}{{{P_1}}}$

Substitute $400\,{\rm{K}}$ for ${T_1}$, $1.00\,{\rm{atm}}$ for ${P_1}$, and $2.00\,{\rm{atm}}$ for ${P_2}$.

$\begin{array}{c}\\{T_2} = \frac{{\left( {2.00\,{\rm{atm}}} \right)\left( {400\,{\rm{K}}} \right)}}{{\left( {1.00\,{\rm{atm}}} \right)}}\\\\ = 800\,{\rm{K}}\\\end{array}$

Hence, the correct options is $800\,{\rm{K}}$.

(2)

The expression for the ideal gas law before leakage of gas is,

${P_2}V = {n_1}R{T_2}$ …… (1)

Rearrange the above in terms of its constant.

$\frac{V}{{R{n_1}}} = \frac{{{T_2}}}{{{P_2}}}$ …… (4)

The expression for the ideal gas law after leakage of gas is,

${P_3}V = \left( {\frac{{{n_1}}}{2}} \right)R{T_3}$ …... (2)

Rearrange the above in terms of its constant.

$\frac{V}{{R{n_1}}} = \frac{{{T_3}}}{{2{P_3}}}$ ……. (5)

Equate the equation (4) and (5).

$\frac{{{T_2}}}{{{P_2}}} = \frac{{{T_3}}}{{2{P_3}}}$

Rearrange the above equation to get the temperature of the gas after leakage of gas.

${T_3} = \frac{{2{P_3}{T_2}}}{{{P_2}}}$

Substitute $2.00\,{\rm{atm}}$for ${P_2}$, $1.00\,{\rm{atm}}$ for ${P_3}$, and $800\,{\rm{K}}$ for ${T_2}$.

$\begin{array}{c}\\{T_3} = \frac{{2\left( {1.00\,{\rm{atm}}} \right)\left( {800\,{\rm{K}}} \right)}}{{\left( {2.00\,{\rm{atm}}} \right)}}\\\\ = 800\,{\rm{K}}\\\end{array}$

Ans: Part 1

The final temperature of the gas is ${\bf{800}}\,{\bf{K}}$.

Part 2

The temperature of the gas after leakage of gas is ${\bf{800}}\,{\bf{K}}$.