A cylindrical flask is fitted with an airtight piston that is free to slide up and down. A mass rests on the top of the piston. The initial temperature of the system is 313 K and the pressure of the gas is held constant at 137 kPa. The temperature is now increased until the height of the piston rises from 23.4 cm to 27.0 cm. What is the final temperature of the gas?

T2 = 361.15 K

Explanation:

Since pressure is held constant, then we'll use Charles law

Charles law gives that V1/T1 = V2/T2

We are given;

Initial temperature; T1 = 313 K

Initial height; h1 = 23.4 cm

Final height; h2 = 27.0 cm

Now, formula for volume of a cylinder is V = πr²h

V1 = πr²h1 and V2 = πr²h2

Now, since V1/T1 = V2/T2 and we are looking for T2, let's make T2 subject;

T2 = (V2 × T1)/V1

Plugging in the relevant values gives;

T2 = (πr²h2 × 313)/πr²h1

πr² will cancel out to give;

T2 = (h2 × 313)/h1

Putting 23.4 for h1 and 27 for h2 gives;

T2 = (27 × 313)/23.4

T2 = 361.15 K

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm

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