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3 years ago

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A mass of 0.1 kg of helium fills a 0.2 m3 rigid tank at 350 kPa. The vessel is heated until the pressure is 700 kPa. Calculate t

he a) the temperature change of helium [deg. C], and b) the total amount of heat required for this process [kJ].

1 answer:

QveST [7]3 years ago

3 0

Explanation:

(a) First, we will calculate the number of moles as follows.

No. of moles = $\frac{\text{mass}}{\text{molar mass}}$

Molar mass of helium is 4 g/mol and mass is given as 0.1 kg or 100 g (as 1 kg = 1000 g).

Putting the given values into the above formula as follows.

No. of moles = $\frac{\text{mass}}{\text{molar mass}}$

= $\frac{\text{100 g}}{4 g/mol}$

= 25 mol

According to the ideal gas equation,

PV = nRT

or, $(P_{2} - P_{1})V = nR (T_{2} - T_{1})$

$(6.90 atm - 3.45 atm) \times 200 L = 25 \times 0.0821 L atm/mol K \Delta T$

$\Delta T$ = 336.17 K

Hence, temperature change will be 336.17 K.

(b) The total amount of heat required for this process will be calculated as follows.

q = $mC \Delta T$

= $100 g \times 5.193 J/g K \times 336.17 K$

= 174573.081 J/K

or, = 174.57 kJ/K (as 1 kJ = 1000 J)

Therefore, the amount of total heat required is 174.57 kJ/K.

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It decomposes in a first-order reaction with a rate constant of 14 s–1. how long would it take for an initial concentration of 0

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$rate = \frac{-(C_{new} - C_{old})}{t}$

where C are the concentration values and t is the time taken for it to decompose.

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$t = 0.003 s$

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Answer: 0.003 s

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3 years ago

Given that an atom of a semiconductor has a diameter of 2.18 å , what is the maximum number of moles that fit in the channel of

andrezito [222]

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3 years ago

A 7600 kg rocket blasts off vertically from the launch pad with a constant upward acceleration of 2.35 m/s2 and feels no appreci

ollegr [7]

Answer:

a) The rocket reaches a maximum height of 737.577 meters.

b) The rocket will come crashing down approximately 17.655 seconds after engine failure.

Explanation:

a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.

1st Stage - Engine

Given that initial velocity, acceleration and travelled distance are known, we determine final velocity ( $v$ ), measured in meters per second, by using this kinematic equation:

$v = \sqrt{v_{o}^{2} +2\cdot a\cdot \Delta s}$ (1)

Where:

$a$ - Acceleration, measured in meters per square second.

$\Delta s$ - Travelled distance, measured in meters.

$v_{o}$ - Initial velocity, measured in meters per second.

If we know that $v_{o} = 0\,\frac{m}{s}$ , $a = 2.35\,\frac{m}{s^{2}}$ and $\Delta s = 595\,m$ , the final velocity of the rocket is:

$v = \sqrt{\left(0\,\frac{m}{s} \right)^{2}+2\cdot \left(2.35\,\frac{m}{s^{2}} \right)\cdot (595\,m)}$

$v\approx 52.882\,\frac{m}{s}$

The time associated with this launch ( $t$ ), measured in seconds, is:

$t = \frac{v-v_{o}}{a}$

$t = \frac{52.882\,\frac{m}{s}-0\,\frac{m}{s}}{2.35\,\frac{m}{s} }$

$t = 22.503\,s$

2nd Stage - Gravity

The rocket reaches its maximum height when final velocity is zero:

$v^{2} = v_{o}^{2} + 2\cdot a\cdot (s-s_{o})$ (2)

Where:

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

If we know that $v_{o} = 52.882\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ , $a = -9.807\,\frac{m}{s^{2}}$ and $s_{o} = 595\,m$ , then the maximum height reached by the rocket is:

$v^{2} -v_{o}^{2} = 2\cdot a\cdot (s-s_{o})$

$s-s_{o} = \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = s_{o} + \frac{v^{2}-v_{o}^{2}}{2\cdot a}$

$s = 595\,m + \frac{\left(0\,\frac{m}{s} \right)^{2}-\left(52.882\,\frac{m}{s} \right)^{2}}{2\cdot \left(-9.807\,\frac{m}{s^{2}} \right)}$

$s = 737.577\,m$

The rocket reaches a maximum height of 737.577 meters.

b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:

$s = s_{o} + v_{o}\cdot t +\frac{1}{2}\cdot a \cdot t^{2}$ (2)

Where:

$s_{o}$ - Initial height, measured in meters.

$s$ - Final height, measured in meters.

$v_{o}$ - Initial speed, measured in meters per second.

$a$ - Gravitational acceleration, measured in meters per square second.

$t$ - Time, measured in seconds.

If we know that $s_{o} = 595\,m$ , $v_{o} = 52.882\,\frac{m}{s}$ , $s = 0\,m$ and $a = -9.807\,\frac{m}{s^{2}}$ , then the time needed by the rocket is:

$0\,m = 595\,m + \left(52.882\,\frac{m}{s} \right)\cdot t + \frac{1}{2}\cdot \left(-9.807\,\frac{m}{s^{2}} \right)\cdot t^{2}$

$-4.904\cdot t^{2}+52.882\cdot t +595 = 0$

Then, we solve this polynomial by Quadratic Formula:

$t_{1}\approx 17.655\,s$ , $t_{2} \approx -6.872\,s$

Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.

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2 years ago

Light moves at a speed of around 1 million miles per hour<br>O<br>True<br>False

AURORKA [14]

Answer:

False

Explanation:

In miles per hour, light speed is about 670,616,629 mph

5 0

3 years ago

Read 2 more answers

During a free fall Swati was accelerating at -9.8m/s2. After 120 seconds how far did she travel? Use the formula =1/2 *

marta [7]

Distance fallen = 1/2 ( V initial + V final ) *t
We know
a = -9.8 m/s2
t=120s

To find distance fallen, we need to find V final
Use the equation
V final = V initial + a*t
Substitute known values
V final = 0 + (-9.8)(120)
V final = -1176 m/s

Then plug known values to distance fallen equation
Distance fallen = 1/2 ( 0 + 1176 )(120)
= 1/2(1776)(120)
=106,560 m

This way plugging into distance equation is actually the long way. A faster way is to plug the values into
Distance fallen = V initial * t + 1/2(a*t)
We won't need to find V final using another equation.

But anyways, good luck!

4 0

3 years ago