Chemistry, 11.07.2019 00:30 stella013108

6. what is the pressure of 0.60 moles of a gas if its volume is 10.0 liters at 35.0°c?

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Answers

The pressure inside the cylinder will be 1.5 atm. The pressure can be calculated using the Ideal Gas Equation which relates the pressure, volume, temperature, and amount of gas to each other.

Further Explanation

The Ideal Gas Equation is:

PV = nRT

where:

P - pressure (in atm)

V - volume (in L)

n - amount of gas (in moles)

R - universal gas constant $0.08206 \frac{L-atm}{mol-K}$

T - temperature (in K)

In the problem, we are given the values:

P = ?

V = 10.0 L (3 significant figures)

n = 0.60 moles ( 2 significant figures)

T = 35.0 degrees Celsius (3 significant figures)

We need to convert the temperature to Kelvin before we can use the Ideal Gas Equation. The formula to convert from degree Celsius to Kelvin is:

$Temperature \ in \ Kelvin = Temperature\ in \ Celsius \ + \ 273.15$

Therefore, for this problem,

$Temperature\ in \ K = 35.0 +273.15\\Temperature\ in \ K = 308.15$

Solving for P using the Ideal Gas Equation:

$P = \frac{nRT}{V} \\P = \frac{(0.60\ mol) \ (0.08206 \ \frac{L-atm}{mol - K}) \ (308.15\ K) \ }{10.0 \ L} \\P = 1.51\ atm$

The least number of significant figures is 2, therefore, the final answer must have only 2 significant figures.

Learn More

1. Learn more about Boyle's Law

2. Learn more about Charles' Law

3. Learn more about Avogadro's Law

Keywords: Ideal Gas Law, Gas Pressure

Answer from: Quest

Work done = force ×distancebut force = mass ×gravityand gravity = 10m\s force = 2×10 = 20 work done = 20×3 = 60 joules

Answer from: Quest

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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