Question

chemistry—unit 2 worksheet 4
on each of the problems below, start with the given p, v, t, or n; then make a decision as to how a change in p, v, t, or n will affect the starting quantity, and then multiply by the appropriate factor. draw particle diagrams of the initial and final conditions.

1. a sample of gas occupies 150 ml at 25 °c. what is its volume when the temperature is increased to 50 °c? (p and n = constant)

	p	t	v	n
final	-	323k	?	-
effect	-	↑	↓	-

150ml×\frac{298k}{323k}=138 ml

1. the pressure in a bicycle tire is 105 psi at 25 °c in fresno. you take the bicycle up to huntington, where the temperature is - 5 °c. what is the pressure in the tire? (v and n = constant)

	p	t	v	n
final	?	268k	-	-
effect	↓	↓	-	-

105psi×\frac{268k}{298k}=94.9 psi

1. what would be the new pressure if 250 cm³ of gas at standard pressure is compressed to a volume of 150 cm³? (t and n = constant)

	p	t	v	n
final	-	-	-	-
effect	-	-	-	-

Explanation:

Step1: Identify the gas - law formula

For problem (1), since $P$ and $n$ are constant, we use Charles's law $\frac{V_1}{T_1}=\frac{V_2}{T_2}$. First, convert temperatures to Kelvin. Initial temperature $T_1 = 25^{\circ}C+273 = 298K$, final temperature $T_2=50^{\circ}C + 273=323K$, and initial volume $V_1 = 150mL$.

Step2: Solve for $V_2$

From $\frac{V_1}{T_1}=\frac{V_2}{T_2}$, we can re - arrange to $V_2=\frac{V_1T_2}{T_1}$. Substitute $V_1 = 150mL$, $T_1 = 298K$, and $T_2 = 323K$ into the formula: $V_2=150\times\frac{323}{298}\approx163mL$.

For problem (2), since $V$ and $n$ are constant, we use Gay - Lussac's law $\frac{P_1}{T_1}=\frac{P_2}{T_2}$. Initial temperature $T_1 = 25^{\circ}C+273 = 298K$, final temperature $T_2=- 5^{\circ}C+273 = 268K$, and initial pressure $P_1 = 105psi$.

Step3: Solve for $P_2$

From $\frac{P_1}{T_1}=\frac{P_2}{T_2}$, we re - arrange to $P_2=\frac{P_1T_2}{T_1}$. Substitute $P_1 = 105psi$, $T_1 = 298K$, and $T_2 = 268K$ into the formula: $P_2=105\times\frac{268}{298}\approx94.9psi$.

For problem (3), since $T$ and $n$ are constant, we use Boyle's law $P_1V_1 = P_2V_2$. Standard pressure $P_1 = 1atm$, initial volume $V_1 = 250cm^3$, and final volume $V_2 = 150cm^3$.

Step4: Solve for $P_2$

From $P_1V_1 = P_2V_2$, we re - arrange to $P_2=\frac{P_1V_1}{V_2}$. Substitute $P_1 = 1atm$, $V_1 = 250cm^3$, and $V_2 = 150cm^3$ into the formula: $P_2=\frac{1\times250}{150}=\frac{5}{3}\approx1.67atm$.

Answer:

(1) The new volume is approximately $163mL$.
(2) The new pressure is approximately $94.9psi$.
(3) The new pressure is approximately $1.67atm$.