Chapter 5, Problem 18P

Interpretation Introduction

Interpretation:

The following table should be completed:

$V_1$	$T_1$	$P_1$	$V_2$	$T_2$	$P_2$
$546\text{L}$	${43}^{\circ }\text{C}$	$6.5\text{atm}$		${65}^{\circ }\text{C}$	$1.9\text{atm}$
$43\text{mL}$	$-{56}^{\circ }\text{C}$	$865\text{torr}$		${43}^{\circ }\text{C}$	$1.5\text{atm}$
$4.2\text{L}$	$234\text{K}$	$0.87\text{atm}$	$3.2\text{L}$	${29}^{\circ }\text{C}$
$1.3\text{L}$	${25}^{\circ }\text{C}$	$740\text{mmHg}$		$0^{\circ }\text{C}$	$1.0\text{atm}$

Answer to Problem 18P

$V_1$	$T_1$	$P_1$	$V_2$	$T_2$	$P_2$
$546\text{L}$	${43}^{\circ }\text{C}$	$6.5\text{atm}$	$1997.9\text{L}$	${65}^{\circ }\text{C}$	$1.9\text{atm}$
$43\text{mL}$	$-{56}^{\circ }\text{C}$	$865\text{torr}$	$47.16\text{mL}$	${43}^{\circ }\text{C}$	$1.5\text{atm}$
$4.2\text{L}$	$234\text{K}$	$0.87\text{atm}$	$3.2\text{L}$	${29}^{\circ }\text{C}$	$1.47\text{atm}$
$1.3\text{L}$	${25}^{\circ }\text{C}$	$740\text{mmHg}$	$1.15\text{L}$	$0^{\circ }\text{C}$	$1.0\text{atm}$

Explanation of Solution

Given information:

Given the below table.

$V_1$	$T_1$	$P_1$	$V_2$	$T_2$	$P_2$
$546\text{l}$	${43}^{\circ }\text{C}$	$6.5\text{a}\text{t}\text{m}$		${65}^{\circ }\text{C}$	$1.9\text{a}\text{t}\text{m}$
$43\text{m}\text{L}$	$-{56}^{\circ }\text{C}$	$865t\text{O}rr$		${43}^{\circ }\text{C}$	$1.5\text{a}\text{t}\text{m}$
$4.2\text{l}$	$234K$	$0.87\text{a}\text{t}\text{m}$	$3.2\text{l}$	${29}^{\circ }\text{C}$
$1.3\text{l}$	${25}^{\circ }\text{C}$	$740mm\text{H}g$		$0^{\circ }\text{C}$	$1.0\text{a}\text{t}\text{m}$

Concept Introduction:

The new volume, temperature or pressure of the gas at the varying temperature, pressures and volumes can be calculated using combined gas law, which states the relationship between pressure, volume and temperature of the gas.

In combined gas law, the three laws of the gas are combined. Mathematically, it is given as.

$\frac{PV}{T}=const\text{a}nt=k$

We two different sets of temperature and pressure of the gas is considered, the above equation becomes as follows:

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

where $P_1$, $V_1$ and $T_1$ are initial pressure, volume and temperature of the gas, while $P_2$, $V_2$ and $T_2$ are final pressure, volume and temperature of the gas.

One quantity can be calculated, we rest are known to us.

The unknown value for each row of information can be calculated using combined gas law.

When $V_1=546\text{l}$, $T_1={43}^{\circ }\text{C}=316.15K$, $P_1=6.5\text{a}\text{t}\text{m}$, $T_2={65}^{\circ }\text{C}=338.15$, $P_2=1.9\text{a}\text{t}\text{m}$

We need to calculate, $V_2$, which is given as follows:

$\frac{6.5\text{atm}\times 546\text{L}}{316.15\text{K}}=\frac{1.9\text{atm}\times V_2}{338.15\text{K}}$

$V_2=\frac{6.\text{5atm}\times 546\text{L}\times 338.15\text{K}}{316.15\text{K}\times 1.9\text{atm}}$

$V_2=1997.9\text{L}$

When $V_1=43\text{mL}$, $T_1=-{56}^{\circ }\text{C}=217.15\text{K}$, $P_1=865\text{torr}=1.13\text{atm}$, $T_2={43}^{\circ }\text{C}=316.15\text{ K}$, $P_2=1.\text{5atm}$

We need to calculate, $V_2$, which is given as follows:

$\frac{1.13\text{atm}\times 43\text{L}}{217.15\text{K}}=\frac{1.\text{5atm}\times V_2}{316.1\text{5K}}$

$V_2=\frac{1.13\text{atm}\times 43\text{L}\times 316.15\text{K}}{217.1\text{5K}\times 1.\text{5atm}}$

$V_2=47.1\text{6mL}$

When $V_1=4.2\text{L}$, $T_1=234\text{K}$, $P_1=0.87\text{atm}$, $V_2=3.2\text{L}$, $T_2={29}^{\circ }\text{C}=302.15\text{ K}$

We need to calculate, $P_2$, which is given as follows:

$\frac{0.87\text{atm}\times 4.2\text{L}}{234\text{K}}=\frac{P_2\times 3.2\text{L}}{302.15\text{K}}$

$P_2=\frac{0.87\text{atm}\times 4.2\text{L}\times 302.15\text{K}}{234\text{K}\times 3.2\text{L}}$

$P_2=1.4\text{7atm}$

When $V_1=1.3\text{L}$, $T_1={25}^{\circ }\text{C}=298.15\text{K}$, $P_1=740\text{mmH}g=0.97\text{atm}$, $T_2=0^{\circ }\text{C}=273.15\text{ K}$, $P_2=1.\text{0atm}$

We need to calculate, $V_2$, which is given as follows:

$\frac{0.97\text{atm}\times 1.3\text{L}}{298.15\text{K}}=\frac{1.0\text{a}\text{t}\text{m}\times V_2}{273.15K}$

$V_2=\frac{0.97\text{atm}\times 1.3\text{L}\times 273.15\mathrm{K}}{298.15\text{K}\times 1.0\text{atm}}$

$V_2=1.15\text{L}$.

Conclusion

Thus, the complete table will be as follows:

$V_1$	$T_1$	$P_1$	$V_2$	$T_2$	$P_2$
$546\text{L}$	${43}^{\circ }\text{C}$	$6.5\text{atm}$	$1997.9\text{L}$	${65}^{\circ }\text{C}$	$1.9\text{atm}$
$43\text{mL}$	$-{56}^{\circ }\text{C}$	$865\text{torr}$	$47.16\text{mL}$	${43}^{\circ }\text{C}$	$1.5\text{atm}$
$4.2\text{L}$	$234\text{K}$	$0.87\text{atm}$	$3.2\text{L}$	${29}^{\circ }\text{C}$	$1.47\text{atm}$
$1.3\text{L}$	${25}^{\circ }\text{C}$	$740\text{mmHg}$	$1.15\text{L}$	$0^{\circ }\text{C}$	$1.0\text{atm}$