A gas occupies 12.3 L at 825.7 mmHg. What will the pressure be when the volume is 75 L?

820 cL

Explanation:

Boyle's Law

Relationship between the pressure and volume of a gas when the temperature is constant  (if we asume the temperature is constant)

It was discovered by Robert Boyle in 1662. Edme Mariotte also came to the same conclusion as Boyle, but he did not publish his works until 1676. This is the reason why in many books we find this law under the name of Boyle and Mariotte's Law.

Boyle's law states that the pressure of a gas in a closed container is inversely proportional to the volume of the container, when the temperature is constant.

As the volume increases, the particles (atoms or molecules) of the gas take longer to reach the walls of the container and therefore collide less times per unit of time against them. This means that the pressure will be lower since this represents the frequency of gas shocks against the walls.

When the volume decreases the distance that the particles have to travel is smaller and therefore more shocks occur in each unit of time: the pressure increases.

What Boyle discovered is that if the amount of gas and the temperature remain constant, the product of pressure by volume always has the same value.

As we have seen, the mathematical expression of this law is:

P * V = K

Where:

P: pressure

V: volumne

K: constant

Suppose we have a certain volume of gas V1 that is at a pressure P1 at the beginning of the experiment. If we vary the volume of gas to a new V2 value, then the pressure will change to P2, and the following will be true:

P1 * V1 = P2 * V2

Calculation

P1 = 40 mmHg = 0.0526 atm

P2 = 60 mmHg = 0.0789 atm

V1 = 12.3 L

V2 = ?

V2 = (P1 * V1) / P2

V2 = (0.0526 atm * 12.3 L) / 0.0789 atm

V2 = 8.2 L = 820 cL

when 25.0 ml of 0.700 mol/l naoh was mixed in a calorimeter with 25.0 ml of 0.700 mol/l hcl, both initially at 20.0 °c, the temperature increased to 22.1 °c. the heat capacity of the calorimeter is 279 j/°c.

the equation for the reaction is:

naoh + hcl → nacl + h₂o

moles of hcl = 0.0250 l hcl ×

0.700

mol hcl

1 l hcl = 0.0175 mol hcl

volume of solution = (25.0 + 25.0) ml = 50.0 ml

mass of solution = 50.0 ml soln ×

1.00

ml soln

= 50.0 g soln

δt=t2-t1

= (22.1 – 20.0) °c = 2.1 °c

the heats involved are

heat from neutralization + heat to warm solution + heat to warm calorimeter = 0

q1+q2+q3=0

nδh+mcδt+cδt=0

0.0175 mol ×

δh

+ 50.0 g × 4.184 j·g⁻¹°c⁻¹ × 2.1 °c + 279 j°c⁻¹ × 2.1 °c = 0

0.0175 mol ×

δh

+ 439.32 j + 585.9 j = 0

0.0175 mol ×

δh

= -1025.22 j

δh=1025.22j0.0175 mol= -58 600 j/mol = -58.6 kj/mol