Think I got the correct answer but I could be way wrong. I found the compression

Question

Think I got the correct answer but I could be way wrong.

I found the compression of the spring to be 214 cm.

NAME DATE Physics 251I Physies 251 Homework 43 Iomework 43 cylinder, oriented horizontally, in an ordinary laboratory has a moveable piston attached to a spring. The cylinder's cross-sectional area is 10 cm2 and it contains 0.040 mol of a polyatomic ideal gas. The constant of the spring is 1500 N/m. At 20 °C the spring is ncither compressed nor stretched. How far is the spring compressed (the spring is sitting between the piston and a wall) if the gas tempcrature is raised to 100 °C, Either find the value of the spring compression or explain why only an approximate value could be found. In either case show all work and explain all the steps

Explanation / Answer

the initial volume can be found:

PV=nRT

(101,300 Pa)(V1)=(0.040mol)(8.31)(293 K)

V1= 0.0009614 m^3

P1V1/T1 = P2V2/T2

P2= 101300 Pa + (1500N/m)x/0.001 m^2 and

V2= V1 + Ax where A is the area of the end of the cylinder

V2= 0.0009614 m^3 + (0.001m^2)x

(101300Pa)(0.0009614 m^3)/(293K) = (101300 Pa + 1500x/0.001 m^2)(0.0009614 m^3 + 0.001x)(1/373K)

0.3324 = (101300 + 1500000x) (0.0009614 + 0.001x) (1/373)

0.3324 = (97.39 + 1442,1x + 101.3x + 1500x^2) (1/373)

0.3324 = 4.02x^2 +4.138x + 0.2611

4.02x^2 + 4.138x -0,0713 =0

x= 1.75766 m = 175,77cm

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