Question

a sample of co₂ gas is in a 755 ml container and has a pressure of 585 mmhg. if the volume of the container is decreased to 482 ml at constant temperature, what is the new pressure? identify the variables from the problem below. enter the numeric value on the right and use the drop down menu to choose the correct unit for the variable. for the unknown variable enter a question mark on the left.

p₁=
v₁=
p₂=
v₂=

calculate the value of the unknown variable. include the calculated value on the left and use the drop - down menu to choose the correct unit.

what is the volume of a sample of argon gas, ar, at 744 torr and 55°c if it occupies 13.3 l at 55°c and 578 torr?

568 l
0.188 l
10.3 l
17.1 l

Explanation:

Step1: Identify the variables

According to the problem, \(P_1 = 585\) mmHg, \(V_1=755\) mL, \(V_2 = 482\) mL, and \(P_2\) is unknown.

Step2: Apply Boyle's Law (\(P_1V_1=P_2V_2\))

$$P_2=\frac{P_1V_1}{V_2}$$

Substitute the values: \(P_1 = 585\), \(V_1 = 755\), \(V_2=482\)

$$P_2=\frac{585\times755}{482}$$

$$P_2=\frac{441675}{482}\approx916$$

mmHg

For the argon gas problem:

Step1: Identify the variables

Using Boyle's Law (\(P_1V_1 = P_2V_2\)), where \(P_1 = 578\) torr, \(V_1=13.3\) L, \(P_2 = 744\) torr

Step2: Solve for \(V_2\)

$$V_2=\frac{P_1V_1}{P_2}$$

Substitute the values: \(P_1 = 578\), \(V_1 = 13.3\), \(P_2=744\)

$$V_2=\frac{578\times13.3}{744}$$

$$V_2=\frac{7687.4}{744}\approx10.3$$

L

Answer:

For the first part:
\(P_1 = 585\) (unit: mmHg), \(V_1 = 755\) (unit: mL), \(P_2=\)? (unit: mmHg), \(V_2 = 482\) (unit: mL), calculated \(P_2\approx916\) mmHg.
For the argon gas problem: The answer is \(10.3\) L (corresponding to the third option).