Question

1. if i have 17 liters of gas at a temperature of 340 k and a pressure of 88.89 atm, what will be the pressure of the gas if i raise the temperature to 367 k and decrease the volume to 12 liters?

p₁= ____ p₂= ____
v₁= ____ v₂= ____
t₁= ____ t₂= ____
\\( p_2 = \frac{p_1 v_1 t_2}{t_1 v_2} \\)

1. i have an unknown volume of gas at a pressure of 0.5 atm and a temperature of 325 k. if i raise the pressure to 1.2 atm, decrease the temperature to 320 k, and measure the final volume to be 48 liters, what was the initial volume of the gas?

p₁= ____ p₂= ____
v₁= ____ v₂= ____
t₁= ____ t₂= ____
\\( p_2 = \frac{p_1 v_1 t_2}{t_1 v_2} \\)

1. if i have 2.9 l of gas at a pressure of 5.0 atm and a temperature of 50.0 c, what will be the temperature of the gas if i decrease the volume of the gas to 2.4 l and decrease the pressure to 3.0 atm?

p₁= ____ p₂= ____
v₁= ____ v₂= ____
t₁= ____ t₂= ____
\\( t_2 = \frac{p_2 v_2 t_1}{p_1 v_1} \\)

1. i have an unknown volume of gas held at a temperature of 115 k in a container with a pressure of 60.0 atm. if increasing the temperature to 225 k and decreasing the pressure to 30.0 atm causes the volume of the gas to be 29 liters, how many liters of gas did i start with?

p₁= ____ p₂= ____
v₁= ____ v₂= ____
t₁= ____ t₂= ____
\\( v_1 = \frac{p_2 v_2 t_1}{t_2 p_1} \\)

1. a gas occupies a volume of 5.0 l at a pressure of 2.5 atm. after conditions change, the gas has a pressure of 1.8 atm, a volume of 7.5 l, and a temperature of 360 k. what was the initial temperature?

p₁= ____ p₂= ____
v₁= ____ v₂= ____
t₁= ____ t₂= ____
\\( t_1 = \frac{p_1 v_1 t_2}{p_2 v_2} \\)

Explanation:

Problem 5

Step1: Identify given values

\( P_1 = 88.89 \, \text{atm} \), \( V_1 = 17 \, \text{L} \), \( T_1 = 340 \, \text{K} \), \( V_2 = 12 \, \text{L} \), \( T_2 = 367 \, \text{K} \)

Step2: Substitute into formula \( P_2 = \frac{P_1 V_1 T_2}{T_1 V_2} \)

\( P_2 = \frac{88.89 \times 17 \times 367}{340 \times 12} \)
Calculate numerator: \( 88.89 \times 17 = 1511.13 \); \( 1511.13 \times 367 \approx 554584.71 \)
Calculate denominator: \( 340 \times 12 = 4080 \)
\( P_2 = \frac{554584.71}{4080} \approx 135.93 \, \text{atm} \)

Step1: Identify given values

\( P_1 = 0.5 \, \text{atm} \), \( P_2 = 1.2 \, \text{atm} \), \( T_1 = 325 \, \text{K} \), \( T_2 = 320 \, \text{K} \), \( V_2 = 48 \, \text{L} \)

Step2: Rearrange formula \( P_2 = \frac{P_1 V_1 T_2}{T_1 V_2} \) to solve for \( V_1 \): \( V_1 = \frac{P_2 V_2 T_1}{P_1 T_2} \)

Substitute values: \( V_1 = \frac{1.2 \times 48 \times 325}{0.5 \times 320} \)
Calculate numerator: \( 1.2 \times 48 = 57.6 \); \( 57.6 \times 325 = 18720 \)
Calculate denominator: \( 0.5 \times 320 = 160 \)
\( V_1 = \frac{18720}{160} = 117 \, \text{L} \)

Step1: Convert temperature to Kelvin: \( T_1 = 50.0 + 273.15 = 323.15 \, \text{K} \)

Given \( P_1 = 5.0 \, \text{atm} \), \( V_1 = 2.9 \, \text{L} \), \( P_2 = 3.0 \, \text{atm} \), \( V_2 = 2.4 \, \text{L} \)

Step2: Substitute into formula \( T_2 = \frac{P_2 V_2 T_1}{P_1 V_1} \)

\( T_2 = \frac{3.0 \times 2.4 \times 323.15}{5.0 \times 2.9} \)
Calculate numerator: \( 3.0 \times 2.4 = 7.2 \); \( 7.2 \times 323.15 \approx 2326.68 \)
Calculate denominator: \( 5.0 \times 2.9 = 14.5 \)
\( T_2 = \frac{2326.68}{14.5} \approx 160.46 \, \text{K} \) (or \( -112.69^\circ \text{C} \))

Answer:

\( P_2 \approx 135.93 \, \text{atm} \) (and \( P_1 = 88.89 \, \text{atm} \), \( V_1 = 17 \, \text{L} \), \( V_2 = 12 \, \text{L} \), \( T_1 = 340 \, \text{K} \), \( T_2 = 367 \, \text{K} \))

Problem 6