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TiliK225 [7]
2 years ago
9

The change in internal energy for the combustion of 1.0 mol of octane at a pressure of 1.0 atm is -5084.1 kj . f the change in e

nthalpy is -5074.3 kj , how much work is done during the combustion
Chemistry
1 answer:
TEA [102]2 years ago
6 0

<u>Given:</u>

Change in internal energy = ΔU = -5084.1 kJ

Change in enthalpy = ΔH = -5074.3 kJ

<u>To determine:</u>

The work done, W

<u>Explanation:</u>

Based on the first law of thermodynamics,

ΔH = ΔU + PΔV

the work done by a gas is given as:

W = -PΔV

Therefore:

ΔH = ΔU - W

W = ΔU-ΔH = -5084.1 -(-5074.3) = -9.8 kJ

Ans: Work done is -9.8 kJ


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1) A car is traveling down the interstate at 37.1 m/s. The driver sees a cop and quickly slows down. If the driver slows to 29.8
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1)

The acceleration of the car is the rate of change of velocity of the car; it can be calculated as:

a=\frac{v-u}{t}

where

u is the initial velocity

v is the final velocity

t is the time taken for the velocity of the car to change from u to v

In this problem, for this car we have:

u = 37.1 m/s

v = 29.8 m/s

t = 3 s

So, the acceleration is:

a=\frac{29.8-37.1}{3}=-2.43 m/s^2

2)

The work done in lifting the box is equal to the potential energy transferred to the box during the process; it is given by:

W=Fd

where

F is the force applied

d is the displacement of the box

Here we have:

F = 87.3 N is the force applied

d = 2.04 m is the displacement of the box

So, the work done to lift the box is:

W=(87.3)(2.04)=178.1 J

3)

The power is the rate of work done per unit time. It is calculated as:

P=\frac{W}{t}

where

W is the work done

t is the time taken to do the work

For the child in this problem, we have:

W = 1250 J is the work done by the child running up the stairs

P = 267 W is the power used

Therefore, re-arranging the equation, we find the time taken:

t=\frac{W}{P}=\frac{1250}{267}=4.68 s

4)

The kinetic energy of an object is the energy possessed by the object due to its motion. Mathematically, it is given by

KE=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

For the rabbit in this problem, we have:

m = 8.642 kg is the mass of the rabbit

KE = 125.6 is its kinetic energy

Solving the formula for v, we find the speed of the rabbit:

v=\sqrt{\frac{2KE}{m}}=\sqrt{\frac{2(125.6)}{8.642}}=5.4 m/s

5)

The efficiency of a machine is the ratio between the energy produced in output by the machine and the work done in input. Mathematically, it is given by

\eta = \frac{E_{out}}{W_{in}}\cdot 100

where

E_{out} is the energy in output

W_{in} is the work in input

For the machine in this problem,

W_{in}=120 J is the work in input

E_{out}=93 J is the energy in output

Therefore, the efficiency of this machine is:

\eta=\frac{93}{120}\cdot 100=77.5\%

6)

During a collision, the total momentum of the system is always conserved before and after the collision. So we can write:

p_i = p_f\\m_1 u_1 + m_2 u_2 =(m_1+m_2)v

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m_1=212 kg is the mass of the first car

u_1=8.00 m/s is the initial velocity of the first car

m_2=196 kg is the mass of the 2nd car

u_2=6.75 m/s is the initial velocity of the 2nd car

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f is the frequency of the wave

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For the wave in the string in this problem we have:

\lambda=0.23 m (wavelength)

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For the blue light in this problem, we have

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d=vt

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