Thermodynamic System and Work Done

Thermodynamics is the branch of science which deals with the measurement of transformation if heat into mechanical work. A thermodynamic system may be the isolated or closed system . work done is different in different process . This not provides us an information on the thermodynamic system and work done.

Summary

Thermodynamics is the branch of science which deals with the measurement of transformation if heat into mechanical work. A thermodynamic system may be the isolated or closed system . work done is different in different process . This not provides us an information on the thermodynamic system and work done.

Things to Remember

Thermodynamics is the branch of science which deals with the measurement of transformation if heat into mechanical work.

A thermodynamic system is in thermal equilibrium if the temperature of all parts of it is same. 

After certain process if the system returns to its initial state is cyclic process is called cyclic process.

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Subjective Questions

Q1:

A point P(6, -4) is translated to the point P'(7, -2). Find the translation vector by which P is translated to P'. With the same translation, what will be the image of A(-4, 6)?


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>P(6, -4) \(\rightarrow\) P' = (x, y)<br>x = 6, y = -4<br>or, P'(7, -2) = (x + a, y + b)<br>or, P'(7, -2) = (6 + a, -4 + b)<br>Equating the corresponding value<br>or, 6 + a = 7 or, -4 + b = -2<br>or, a = 7 - 6 or, b = -2 + 4<br>\(\therefore\) a = 1 \(\therefore\) b = 2<br>Here, The vector is \(\begin{pmatrix} 1 \\ 2 \end{pmatrix}\).<br>Now,<br>or, A(-4, 6) = A'(x + a, y + b)<br>or, A(-4, 6) = A'(-4 + 1, 6 + 2)<br>or, A(-4, 6) = A'(-3, 8)<br>\(\therefore\) The tarnslation of given point is A'(-3, 8).</p>

Q2:

Under a translation, a point A(6, 4) is translated to A'(3, 2). Where will the point Q(2, -4) the translated by the same translation vector?


Type: Short Difficulty: Easy

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Answer: <p>Solution:</p> <p>Here,<br>A(6, 4) = A'(x, y)<br>or, x = 6, y = 4<br>or, A'(3, 2) = (x + a, y + b)<br>or, A'(3, 2) = (6 + a, 4 + b)<br>Equalating the corresponding value<br>or, 6 + a = 3 or, 4 + b = 2<br>or, a = 3 - 6 or, b = 2 - 4<br>\(\therefore\) a = -3 \(\therefore\) b = -2<br>\(\therefore\) The translation vector is \(\begin{pmatrix} -3 \\ -2 \end{pmatrix}\)<br>Now,<br>or, Q(2, -4) = Q'(x + a, y + b)<br>or, Q(2, -4) = Q'(2 - 3, -3 - 2)<br>or, (2, -4) = Q'(-1, -6)<br>\(\therefore\) The image of Q' is (-1, -6)</p>

Q3:

The point G(5, -5) is reflected over the y-axis. What are the coordinates of the resulting point, G ?


translation


Type: Short Difficulty: Easy

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Answer: <p>Reflect the point over the line.</p> <figure><img src="/uploads/Capture_2.JPG" alt="tr" width="196" height="193"></figure><p>Since G is 5 units to the right of the y-axis, G&prime; is 5 units to the left of the y-axis. G&prime; has coordinates (-5,-5).</p> <p></p>

Q4:

The point T(-3, 2) is reflected over the x-axis. What are the coordinates of the resulting point, T′?

f


Type: Short Difficulty: Easy

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Answer: <p>Reflect the point over the line.</p> <figure><img src="/uploads/Capture_4.JPG" alt="f" width="236" height="202"></figure><figure>Since T is 2 units above the x-axis, T&prime; is 2 units below the x-axis. T&prime; has coordinates (-3,-2).</figure>

Q5:

Find the image of the following points when they are translated by the vector, Q(4, -7)
T = \(\begin{pmatrix} -2 \\ 4 \end{pmatrix}\) and T = \(\begin{pmatrix} 3 \\ -1 \end{pmatrix}\)


Type: Short Difficulty: Easy

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Answer: <p>Solution:</p> <p>Here,<br>P(x, y)<img src="/uploads/trans_1.jpg" alt="q" width="48" height="43">P' (x + a, y + b)<br>P(6, 4)<img src="/uploads/mhfj.jpg" alt="g" width="43" height="42">P' (4 + (-2), -7 - 4) = P'(2, -11)<br>P(4, -7)<img src="/uploads/yi.jpg" alt="r" width="49" height="42">P' (4 + 3, -7 - 1) = P' (7, -8)</p>

Q6:

Find the image of the points A(4, 6) and B(-5, 7) when they are translated by the vector T =  \(\begin{pmatrix} 4 \\ -3 \end{pmatrix}\).


Type: Very_short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>If \(\begin{pmatrix} a\\ b\end{pmatrix}\) is the translation vector and P(x, y) is an object then</p> <p>P(x, y) \(\rightarrow\) P'(x+a, y+b)</p> <p>Here, T = \(\begin{pmatrix} 4\\ -3 \end{pmatrix}\)</p> <p>\(\therefore\) A(4, 6) \(\rightarrow\) A'(4+4, 6-3) = A'(8, 3)</p> <p>And B(-5, 7) \(\rightarrow\) B'(-5+4, 7-3) = B'(-1, 4)</p>

Q7:

Under translation, a point P(4, 5) is translated to P'(2, 1). What will be the image of Q(3, -5) by the same translation vector?


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>Here P(4, 5) = P(x, y)</p> <p>\(\therefore\) x = 5, y = 5</p> <p>P'(2, 1) = P'(x+a, y+b)</p> <p>or, P'(2,1) = P'(4+a, 5+b)</p> <p>\(\therefore\) 4+a = 2 and 5+b = 1, equating the corresponding elements.</p> <p> or, a = 2-4 or, b = 1-5</p> <p> or, a = -2 or, b = -4</p> <p>\(\therefore\) Translation vector =\(\begin{pmatrix} a\\ b\end{pmatrix}\) =\(\begin{pmatrix} -2 \\ -4 \end{pmatrix}\)</p> <p>Now, Q(3, -5)\(\rightarrow\) Q'(x+a, y+b) = Q'(3-2, -5-4)</p> <p>= Q'(1, -9)</p> <p>\(\therefore\)The image of Q is Q'(1, -9).</p>

Q8:

The vertices of \(\triangle\)ABC are A(0, 4), B(3, -2) and C(6, 0). Translate \(\triangle\)ABC, 5 units upward. Draw \(\triangle\) and its image on the graph.


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>The vertices of \(\triangle\) are A(0, 4), B(3, -2) and C(6, 0).</p> <p>Translation vector (T) =\(\begin{pmatrix} 0\\5 \end{pmatrix}\)</p> <p>When , T =\(\begin{pmatrix} a\\b \end{pmatrix}\), P(x, y) \(\rightarrow\) P'(x+a, y+a)</p> <p>\(\therefore\) A(0, 4)\(\rightarrow\) A'(0+0, 4+5) = A'(0, 9)</p> <p>B(3, -2)\(\rightarrow\) B'(3+0, -2+5) = B'(3, 3)</p> <p>C(6, 0) \(\rightarrow\) C'(6+0, 0+5) = C'(6, 5)</p> <p>\(\triangle\)ABC and \(\triangle\)A'B'C' are shown on the graph.</p> <figure class="" style="width: 350px;"><img src="/uploads/sadda.JPG" alt="z" width="350" height="367"><figcaption><br></figcaption></figure><p></p> <figure class="" style="width: 350px;"><br><figcaption><br></figcaption></figure><p></p>

Q9:

The vertices of \(\triangle\)ABC are A(2, 5), B(6, -4) and C(8, 0). Translate \(\triangle\)ABC, 3 units upward. Draw \(\triangle\)ABC and its image on the graph.


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>The vertices of \(\triangle\) are A(2, 5), B(6, -4) and C(8, 0).</p> <p>Translation vector (T) =\(\begin{pmatrix} 0\\3 \end{pmatrix}\)</p> <p>When , T =\(\begin{pmatrix} a\\b \end{pmatrix}\), P(x, y) \(\rightarrow\) P'(x+a, y+a)</p> <p>\(\therefore\) A(2, 5) \(\rightarrow\) A'(2+0, 5+3) = A'(2, 8)</p> <p>B(6, -4)\(\rightarrow\) B'(6+0, -4+3) = B'(6, -1)</p> <p>C(8, 0) \(\rightarrow\) C'(8+0, 0+3) = C'(8, 3)</p> <p>\(\triangle\)ABC and \(\triangle\)A'B'C' are shown on the graph.</p> <figure class="" style="width: 350px;"><img src="/uploads/30.JPG" alt="z" width="350" height="387"><figcaption><br></figcaption></figure>

Q10:

The vertices of \(\triangle\)ABC are A(2, 3), B(4, 5) and C(6, 4). Translate \(\triangle\)ABC by the translation vector \(\begin{pmatrix} 2\\3 \end{pmatrix}\). Draw \(\triangle\)ABC and its image on the graph.


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>The vertices of \(\triangle\) are A(2, 3), B(4, 5) and C(6, 4).</p> <p>Translation vector (T) = \(\begin{pmatrix} 2\\3 \end{pmatrix}\)</p> <p>When, T = \(\begin{pmatrix} a \\ b \end{pmatrix}\), then</p> <p>P(x, y) \(\rightarrow\) P'(x+a, y+a)</p> <p>\(\therefore\) A(2,3) \(\rightarrow\) A'(2+2, 3+3) = A'(4, 6)</p> <p>or, B(4, 5) \(\rightarrow\) B'(4+2, 5+3) = B'(6, 8)</p> <p>or, C(6, 4) \(\rightarrow\) C'(6+2, 4+3) = C'(8, 7)</p> <p>\(\triangle\)ABC and \(\triangle\)A'B'C' are shown on the graph.</p> <figure class="" style="width: 350px;"><img src="/uploads/311.JPG" alt="z" width="350" height="386"><figcaption><br></figcaption></figure>

Q11:

The vertices of \(\triangle\)ABC are A(2, 5), B(1, -4) and C(3, 0). Translate \(\triangle\)ABC  by the translation vector \(\begin{pmatrix} 2\\3 \end{pmatrix}\). Draw \(\triangle\) and its image on the graph.


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>The vertices of \(\triangle\) are A(2, 5), B(1, -4) and C(3, 0).</p> <p>Translation vector (T) = \(\begin{pmatrix} 0\\5 \end{pmatrix}\)</p> <p>When, T = \(\begin{pmatrix} a\\b \end{pmatrix}\), P(x, y) \(\rightarrow\) P'(x+a, y+a)</p> <p>\(\therefore\) A(2, 5) \(\rightarrow\) A'(2+2, 5+3) = A'(4, 8)</p> <p>or, B(1, -4) \(\rightarrow\) B'(1+2, -4+3) = B'(3, -1)</p> <p>or, C(3, 0) \(\rightarrow\) C'(3+2, 0+3) = C'(5, 3)</p> <p>\(\triangle\)ABC and \(\triangle\)A'B'C' are shown on the graph.</p> <p></p> <figure class="" style="width: 350px;"><img src="/uploads/czx.JPG" alt="z" width="350" height="367"><figcaption><br></figcaption></figure><p></p>

Q12:

The vertices of \(\triangle\)XYZ are X(3, 1), Y(4, 3) and Z(1, 4). Translate \(\triangle\)XYZ by the translation vector \(\begin{pmatrix} 3\\4 \end{pmatrix}\). Draw \(\triangle\)XYZ and its image on the graph.


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>The vertices of \(\triangle\) are X(3, 1), Y(4, 3) and Z(1, 4).</p> <p>Translation vector (T) = \(\begin{pmatrix} 3\\4 \end{pmatrix}\)</p> <p>When, T = \(\begin{pmatrix} a\\b \end{pmatrix}\), then</p> <p>P(x, y) \(\rightarrow\) P'(x+a, y+a)</p> <p>\(\therefore\) X(3, 1) \(\rightarrow\) X'(3+3, 1+4) = X'(6, 5)</p> <p>or, Y(4, 3) \(\rightarrow\) Y'(4+3, 3+4) = Y'(7, 7)</p> <p>or, Z(1, 4) \(\rightarrow\) Z'(1+3, 4+4) = Z'(4, 8)</p> <p>\(\triangle\)XYZ and \(\triangle\)X'Y'Z' are shown on the graph.</p> <figure class="" style="width: 350px;"><img src="/uploads/saaa1.JPG" alt="z" width="350" height="366"><figcaption><br></figcaption></figure><p></p>

Q13:

The vertices of \(\triangle\)ABC are A(-2, 3), B(3, 4) and C(2, 1). Translate \(\triangle\)ABC by the translation vector \(\begin{pmatrix} 2\\3 \end{pmatrix}\). Draw \(\triangle\) ABC  and its image on the graph.


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>The vertices of \(\triangle\) are A(-2, 3), B(3, 4) and C(2, 1)</p> <p>Translation vector (T) = \(\begin{pmatrix} 2\\3 \end{pmatrix}\)</p> <p>When, T = \(\begin{pmatrix} a\\b \end{pmatrix}\), then</p> <p>P(x, y) \(\rightarrow\) P'(x+a, y+a)</p> <p>\(\therefore\) A(-2,3) \(\rightarrow\) A'(-2+2, 3+3) = A'(0, 6)</p> <p>or, B(3, 4) \(\rightarrow\) B'(3+2, 4+3) = B'(5, 7)</p> <p>or, C(2, 1) \(\rightarrow\) C'(2 +2, 1+3) = C'(4, 4)</p> <p>\(\triangle\) ABC and \(\triangle\) A'B'C' are shown on the graph.</p> <figure class="" style="width: 350px;"><img src="/uploads/1222222222222222222222222.JPG" alt="z" width="350" height="369"><figcaption><br></figcaption></figure>

Q14:

The vertices of \(\triangle\)PQR are P(2, 5), Q(3, -4) and R(4, 1). Translate \(\triangle\)PQR by the translation vector \(\begin{pmatrix} 1\\3 \end{pmatrix}\). Draw \(\triangle\)PQR and its image on the graph.


Type: Long Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>The vertices of \(\triangle\) are P(2, 5), Q(3, -4) and R(4, 1).</p> <p>Translation vector (T) = \(\begin{pmatrix} 1\\3 \end{pmatrix}\)</p> <p>When, T = \(\begin{pmatrix} a\\b \end{pmatrix}\), P(x, y) \(\rightarrow\) P'(x+a, y+a)</p> <p>\(\therefore\) P(2, 5) \(\rightarrow\) P'(2+1, 5+3) = P'(3, 8)</p> <p>or, Q(3, -4) \(\rightarrow\) Q'(3+1, -4+3) = Q'(4, -1)</p> <p>or, R(4, 1) \(\rightarrow\) R'(4+1, 1+3) = R'(5, 4)</p> <p>\(\triangle\)PQR and \(\triangle\)P'Q'R' are shown on the graph.</p> <figure class="" style="width: 350px;"><img src="/uploads/12222222222222222222222221.JPG" alt="xz" width="350" height="371"><figcaption><br></figcaption></figure>

Q15:

A point P(3, -2) is translated to the point P'(4, -1). Find the translation vector by which P is translated to P'. With the same translation, what will be the image of A(-2, 3) ?


Type: Short Difficulty: Easy

Show/Hide Answer
Answer: <p>Solution:</p> <p>P(3, -4) \(\rightarrow\) P' = (x, y)<br>x = 3, y = -4<br>or, P'(4, -1) = (x + a, y + b)<br>or, P'(4, -1) = (3 + a, -4 + b)<br>Equating the corresponding value<br>or, 3 + a = 4 or, -4 + b = -1<br>or, a = 4 - 3 or, b = 4 - 1<br>\(\therefore\) a = 1 \(\therefore\) b = 3<br>Here, The vector is \(\begin{pmatrix} 1 \\ 3\end{pmatrix}\).<br>Now,<br>or, A(-2, 3) = A'(x + a, y + b)<br>or, A(-2, 3) = A'(-2 + 1, 3 + 3)<br>or, A(-2, 3) = A'(-1, 6)<br>\(\therefore\) The tarnslation of given point is A'(-1, 6).</p> <p></p>

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Thermodynamic System and Work Done

Thermodynamic System and Work Done

Introduction

Thermodynamics is the branch of science which deals with the measurement of transformation if heat into mechanical work. That means it mainly describes the inter-relationship between heat and mechanical energy.

Thermodynamic System

It is an assembly of an extremely large number of particles having a certain pressure, volume and temperature. A thermodynamic system can exchange energy with its surrounding by transfer or do mechanical work. The state of a system is the physical condition of system which are described by pressure, volume and temperature. A thermodynamic system may be isolated or closed system.

Thermal Equilibrium

A thermodynamic system is in thermal equilibrium if the temperature of all parts of it is same. So, a system in thermal equilibrium does not exchange heat between its different parts or with surroundings.

Work Done

Let us consider an ideal gas is enclosed in a cylinder fitted with a frictionless moveable piston. Let the pressure exerted by the gas be P and the cross-sectional area of the piston be A. The force exerted by the gas on the piston is F= PA.

Work Done by Gas during Expansion

Work done by a gas during an expansion of dx.
Work done by a gas during an expansion of dx.

When the gas expands, the piston moves out through a small distance dx

and work done by the force

$$dW = F dx = PA dx$$

Since \(A dx = dV\), a small increase in the volume of the gas, the work done by the gas during the expansion.

$$dW = PdV \dots (i)$$

When the volume of the gas changes from V1 to V2, the total work done W is obtained by integrating the equation (i) within the limits V1 to V2.

$$W = \int dW = \int _{V_1}^{V_2} P dV \dots (ii)$$

When the final volume V2 is greater than initial volume V1, then the change in volume V2- V1 is positive. Hence during the expansion of the gas work done by a system is taken as positive.

When the gas is compressed, the final volume V2 is less than the initial volume V1then the change in volume V2- V1 is negative.Hence during the compression of the gas work done by a system is taken as negative.

g
(a) Indicator diagram showing the work done by gas. (b) Work done by gas on expansion at constant pressure.
Indicator Diagram

Let E be a point on the indicator diagram. P and V be the pressure and volume at the point. Let the volume increase by small amount dV at constant P to a point F which is close to E.

From the figure, we have

Area of the small strip \( EFGH = HG \times EH = dV \times P = PdV \) where dV is a small volume represented by HG during which the pressure P = HE remains constant.

\(\therefore\) The area of the strip EFGH = PdV = work done during a small change of volume dV.

Thus, the work done during expansion from the initial state A(P1, V1) to the final state (P2, V2) can be obtained by adding the area of such small stripes

$$W = \text {area of ABCD}$$

Thus work done by a system is numerically equal to the area under PV-diagram. So,

$$W = \int _{V_1}^{V_2} P dV= \text {area of ABCD} $$

When the pressure P remains constant throughout the expansion, the work done by the gas is

$$W = P(V_2 - V_1)$$

Work done during a cyclic process
Work done during a cyclic process

Work Done in a Cyclic Process

After thecertain process, if the system returns to its initial state is cyclic process is called cyclic process.

Let us consider a system expand from Initial stage A to the state B along the path X and the system compress from B to A along the path Y which is shown in the figure.

The work done by system when system expand from A to B is represented by the curve AXB and volume axis that is

$$W_1 = \text {area of AXBB'A'A}$$

The work done on the system when system contract fromBA is given by the area between curve BYA and volume axis i.e.

$$W_2 = -(\text {area of BYAA'B'B})$$

$$\therefore \text {Net work done in the cyclic process is}$$

$$W = W_1 + W_2$$

$$= \text {area of AXBB'A'A} +(\text {-area of BYAA'B'B})$$

$$= + \text {area of AXBYA} $$

Lesson

First Law of Thermodynamics

Subject

Physics

Grade

Grade 11

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