1. Plot these two state points on a pressure (ordinate) - volume (abscissa) plane: at state $1, P_1=60 {Bar}, {V}_1=100 {li}$; at state $2, {p}_2=10 {bar}, {V}_2=700 {li}$. Now join them with a single straight line. (a) What will be the pressure and volume of a third state point located on this line and mid-way between the first two state points? (b) From a right triangle using the straight line as the hypotenuse. What will be the pressure and volume of the state point located at the junction of the two legs of the triangle?

Answers

Answer 1

(a) The pressure and volume of the third state point located midway between the first two state points will be approximately 35 Bar and 400 li, respectively.

(b) The pressure and volume of the state point located at the junction of the two legs of the right triangle will be approximately 40 Bar and 250 li, respectively.

(a) To find the pressure and volume of the third state point, we can use the concept of linear interpolation. Since the two given state points are joined by a straight line, we can determine the pressure and volume at the midpoint by taking the average of the corresponding values of the two points. Thus, the pressure at the third state point is (60 + 10)/2 = 35 Bar, and the volume is (100 + 700)/2 = 400 li.

(b) In a right triangle, the hypotenuse represents the straight line joining the two state points. By using the Pythagorean theorem, we can calculate the length of the hypotenuse, which corresponds to the pressure and volume at the junction of the two legs. The difference in pressure between the two state points is 60 - 10 = 50 Bar, and the difference in volume is 700 - 100 = 600 li. Treating these differences as the legs of a right triangle, we can calculate the hypotenuse length using the theorem. The pressure at the junction point is given by sqrt((40^2) + (50^2)) = 40 Bar, and the volume is sqrt((250^2) + (600^2)) = 250 li.

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Related Questions

The opcode of MOV AX, BX is AX or BX. Select one: True False

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The opcode of the instruction MOV AX, BX is neither AX nor BX, but rather the binary code 10001011. This instruction moves the contents of register BX into register AX in the x86 assembly language.

This can be explained as follows:Opcode is the hexadecimal value of the operation code (op code) used to specify the instruction to be executed in a computer's central processing unit (CPU).It tells the processor what to do with the operands of the instruction (such as AX, BX in this case). The opcode is used to assemble machine code, which the CPU can then execute. In general, opcodes are used to specify operations that can be performed by a processor, such as moving data between registers (MOV), adding numbers (ADD), and comparing data (CMP).The opcode of MOV AX, BX is a two-byte instruction in which the first byte specifies the opcode (8B in hexadecimal) and the second byte specifies the registers involved in the instruction (C3 in hexadecimal). The instruction is encoded as follows: 8B C3. Hence, we can say that the answer is false.

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