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3 years ago

6

A gas is confined to a cylinder under constant atmospheric pressure, as illustrated in the following figure. When the gas underg

oes a particular chemical reaction, it absorbs 829J of heat from its surroundings and has 0.69kJ of P−V work done on it by its surroundings. What is the value of ΔH for this process?

2 answers:

Lyrx [107]3 years ago

8 0

Answer:The value of ΔH for this process is 2209 Joules.

Explanation:

Work done on the system by its surroundings,W = $P\Delta V$ = 0.69 kJ = 690 J

Heat added or absorbed by the system,Q = 829 J

According to first law of thermodynamics:

$\Delta U=Q+W=829 J+690 J=1519 J$

The $\Delta H$ of the process is given by:

$\Delta H=\Delta U+P\Delta V=1519 J+690 J=2209 J$

The value of ΔH for this process is 2209 Joules.

GarryVolchara [31]3 years ago

7 0

From the first law of thermodynamics, we use the equation expressed as:

ΔH = Q + W

where Q is the heat absorbed of the system and W is the work done.

We calculate as follows:

ΔH = Q + W
ΔH = 829 J + 690 J = 1519 J

Hope this answers the question. Have a nice day.

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Explanation:

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$0-u^2=2\left ( -g\right )h$

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3 years ago

Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin.

Sergio039 [100]

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge $q_{1}= 3.8\times10^{-6}\ C$

Second charge $q_{2}=3.2\times10^{-6}\ C$

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

$E_{1}=E_{2}$

$\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}$

$\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}$

$\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}$

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Put the value into the formula

$x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}$

$x=1.69\ m$

Hence, The distance is 1.69 m.

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3 years ago

A wheel with moment of inertia 25 kg. m2 and angular velocity 10 rad/s begins to speed up, with angular acceleration 15 rad/sec2

Pani-rosa [81]

Answer:

(A) Angular speed 40 rad/sec

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We have given t = 2 sec

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3 years ago

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tatyana61 [14]

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$(1 + x)^{a} = 1 + ax$

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