A sample of water in equilibrium with its vapor in a closed 3-L container has a vapor pressure of 23.8 torr at 25°C. The container's volume is increased to 7 L at constant temperature and the liquid/vapor equilibrium is reestablished. What is the vapor pressure?

The vapor pressure of this liquid in a closed container remains the same and as such would be equal to 23.8 torr.

Given the following data:

To determine the final vapor pressure of a sample of water in equilibrium with its vapor:

What is vapor pressure?

Vapor pressure can be defined as the pressure exerted by a vapor in thermodynamic equilibrium with a liquid or solid that is placed in a closed container at a specific temperature.

On a related note, the vapor pressure of a liquid is the pressure exerted by a vapor when its liquid in a closed container evaporates above the liquid sample.

At a constant temperature, the vapor pressure of a liquid in a closed container would always remain the same provided that vapor equilibrium is reached in the container having a new volume.

In this context, the vapor pressure of this liquid in a closed container remains the same and as such would be equal to 23.8 torr.

Read more on vapor pressure here: link

23.8 torr

Explanation:

Vapor pressure of a liquid is simply the pressure of the vapor that arises when the liquid evaporates in a closed container above the liquid sample.

From the question given above, the vapor pressure is 23.8 torr at 25°C. Since the temperature is constant, the vapor will remain the same when equilibrium is reached in the new container as vapor pressure depends on the temperature of the liquid.

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

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