pv diagram practice problems

0000001713 00000 n Physics Q&A Library Heat Engine Practice Problem Problem Statement A piston-cylinder heat engine containing a monatomic ideal gas undergoes the three processes drawn on the p-V diagram below. 0000000879 00000 n 0000000781 00000 n The diagram in Fig. (b) The Gas Is Then Cooled At Constant Volume Until The Pressure Falls To 1.5 Atm. chemical potential) is varied. So the actual shape of the line drawn on a PV diagram for an isothermal process is sometimes called an isotherm and they look like that. Also, the T used is not room temperature, but the temperature given in the problem – the temperature at which the reaction takes place. Phase rule. If you're seeing this message, it means we're having trouble loading external resources on our website. ���!�y¤ �n;000NpϾ�D�Yh:�I>�3�XB���������l����((����� (�����@��E�"~ܜ O�f)��c`T6QU���!˜��wpG��SB��20�q� ɖ 4 endstream endobj 49 0 obj 159 endobj 41 0 obj << /Type /Page /Parent 35 0 R /Resources 42 0 R /Contents 46 0 R /MediaBox [ 0 0 612 792 ] /CropBox [ 0 0 612 792 ] /Rotate 0 >> endobj 42 0 obj << /ProcSet [ /PDF /Text ] /Font << /TT2 44 0 R >> /ExtGState << /GS1 47 0 R >> /ColorSpace << /Cs5 45 0 R >> >> endobj 43 0 obj << /Type /FontDescriptor /Ascent 891 /CapHeight 0 /Descent -216 /Flags 34 /FontBBox [ -568 -307 2028 1007 ] /FontName /TimesNewRoman /ItalicAngle 0 /StemV 0 >> endobj 44 0 obj << /Type /Font /Subtype /TrueType /FirstChar 32 /LastChar 122 /Widths [ 250 0 0 0 0 0 0 0 333 333 0 0 250 333 250 278 500 500 500 500 500 500 500 500 500 0 278 0 0 564 0 444 0 722 0 667 722 611 556 0 722 333 389 722 0 889 722 722 556 0 667 556 611 722 722 944 722 0 611 0 0 0 0 0 0 444 500 444 500 444 333 500 500 278 278 500 278 778 500 500 500 500 333 389 278 500 500 722 500 500 444 ] /Encoding /WinAnsiEncoding /BaseFont /TimesNewRoman /FontDescriptor 43 0 R >> endobj 45 0 obj [ /CalRGB << /WhitePoint [ 0.9505 1 1.089 ] /Gamma [ 2.22221 2.22221 2.22221 ] /Matrix [ 0.4124 0.2126 0.0193 0.3576 0.71519 0.1192 0.1805 0.0722 0.9505 ] >> ] endobj 46 0 obj << /Length 1531 /Filter /FlateDecode >> stream [-11 Points] PRACTICE ANOTHER ASK YOUR TEACHER DETAILS MY NOTES SERCP11 12.1.P.002. 0000003995 00000 n Pressure-volume graphs are used to describe thermodynamic processes — especially for gases. Use the P-V diagram below to answer the following questions 1a) The Net Work for the cyclic process is: a) Zero b) Positive c) Negative d) Cannot tell from the diagram 1b) The processes from states 1 to 2 and 3 to 4 are: a) Isothermal b) Isobaric c) Isochoric d) Isometric

If you're behind a web filter, please make sure that the domains Our mission is to provide a free, world-class education to anyone, anywhere.Khan Academy is a 501(c)(3) nonprofit organization.Reconciling thermodynamic and state definitions of entropyCarnot efficiency 3: Proving that it is the most efficientReconciling thermodynamic and state definitions of entropyCarnot efficiency 3: Proving that it is the most efficientLearn what PV diagrams are and how to use them to find the change in internal energy, work done, and heat. Phase rule. 0000001237 00000 n

So on a PV diagram, an isothermal process is gonna look something like this, it's gonna curve like 1/x and it can be an isothermal expansion if volume increases or an isothermal compression if volume decreases. 0000001526 00000 n If you're seeing this message, it means we're having trouble loading external resources on … Fill in the table below, where C = the number of independent components, P = number of phases and F = degrees of freedom in the following system at equilibrium. Q # 1. What is the total work done by the gas and what is the total heat flow into the gas after completing one cycle? Pure liquid water … System C p. Pure liquid water. Question: 7 F 0 05 2. ΔG=ΔH-TΔS ΔG=[-57.12]-310.5[-.1757] = -2.565 kJ/mol or -2565 J/mol Sketch A PV Diagram And Find The Work Done By The Gas During The Following Stages (a) A Gas Is Expanded From A Volume Of 1.0 L To 5.0 L At A Constant Pressure Of 7.0 Atm. 0000000627 00000 n Fill in the table below, where C = the number of independent components, P = number of phases and F = degrees of freedom in the following system at equilibrium. Phase rule, PT diagram, PV diagram. ��GHi�����{F�܀]M�}�;�s(#�|„ÜF���1�4�s���C���޸�'��I~����&�yn���WX|�*�x�?���Jq�)��2�>�ރdAh��@m�E�e7�z��� 2]�"X�K�Y!b��pA�K���\�"�(2G�`N��j_N�:�N�z|N��pӦI�x���O�gY'{ Practice Problems, Chapters 1 - 4 1. Phase rule, PT diagram, PV diagram. 36 0 obj << /Linearized 1 /O 41 /H [ 965 272 ] /L 23495 /E 4238 /N 5 /T 22657 >> endobj xref 36 14 0000000016 00000 n

Pure liquid water … 0000000817 00000 n System C p. Pure liquid water. trailer << /Size 50 /Info 34 0 R /Root 37 0 R /Prev 22647 /ID[<70a337057a355f9bdafd11bb4dd585e9><70a337057a355f9bdafd11bb4dd585e9>] >> startxref 0 %%EOF 37 0 obj << /Type /Catalog /Pages 35 0 R /Outlines 32 0 R /Threads 38 0 R /OpenAction [ 41 0 R /XYZ null null null ] /PageMode /UseNone >> endobj 38 0 obj [ 39 0 R 40 0 R ] endobj 39 0 obj << /I << /Title (P=800kPa)>> /F 10 0 R >> endobj 40 0 obj << /I << /Title (\(Tank volumes are NOT to scale\))>> /F 2 0 R >> endobj 48 0 obj << /S 108 /O 163 /Filter /FlateDecode /Length 49 0 R >> stream H�b```��,�x�(� However, many useful phase diagrams can be drawn which involve variables other than T and composition. Learn what PV diagrams are and how to use them to find the change in internal energy, work done, and heat. �����^;l�t����}�NQ���d�!�?M��uH��T���ՙ@m=�J��0���;�_�W�4���PWm�sE�!��V�(M��G]���k��]���Q���:55�����ݧ 1�/��-fˁ��aߊ�}rh�=�U�3��3n�+����J��Τ�~�kן�Fu5�aRƷ��������|W4_��W7��B�j�c-�"����v�_�r}��g�;�K" �L�.����>�� ����;��'�#�� �O��V)�dyv�$\�x�q)XuYH���S��N52���Gq75�P�H[���ء�? The gas is initially at room temperature (300 K). Work, heat, and changes in internal energy can also be determined. ]յh��ӫ��v������7J���(K�E�fZ�J4��[�m^A�����/���u�Z),jY>M�|��.�g%���kZ���&��۴�'/�!��F)�m`�W�wͫ�XgD�]DX����u�m�}ۉ�>�����sEȆ|���7�j�E� ?҆�cG�I+��R���1b]�>�&.���QêPu]QKPRh�6>[ﲚ��F��ll��JF��M���Ea��y ��N�;� ��q�'���{���Gq��V�x�Q�i%��H�>�% 0000001391 00000 n

Q # 1.

0000000965 00000 n

0000002209 00000 n

5 shows the phases present at equilibrium in the Fe-Ni-O 2 system at 1200 oC as the equilibrium oxygen partial pressure (i.e. )�݉�j�#�d���Ri�O��i�i��X6D�'&,oo�.��y��L���pwt�� �}#+~�Fs�ny@V�F��T��}0w�u=�na�ˁ��Q�݉VN��sV�k � ݭw� ��t���vs6_�� %PDF-1.2 %����