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As shown in the figure, an enclosed cylindrical container (cross-sectional area: S) is...

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kibauu | eNoter

Posted May 29, 2013 at 2:26 PM via web

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As shown in the figure, an enclosed cylindrical container (cross-sectional area: S) is divided into two sections (A,B)by a piston P. The length of each section is 10 cm. Each section contains a monatomic ideal gas at temperature 0Celcius and pressure 1,0x10^5 Pa (both contain the same type of gas). The gas in B is in contact with a thermostatic bath at temperature 0 Celcius, and doesnt change in temperature. The piston and container do not conduct heat. The area of contact between the piston and the container is tightly sealed and frictionless.

When the gas in A is heated to 57 celcius using a heater, what distance is traveled by P?

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llltkl | College Teacher | Valedictorian

Posted May 29, 2013 at 5:08 PM (Answer #1)

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Let the piston travels x cm towards chamber B to reach equilibrium after the gas in chamber A is heated to 57°C. So, the length of chamber A now increased to (10+x) cm and that of chamber B has been reduced to (10-x) cm. their volumes are obtained by multiplying cross sectional area (S), with length.

As the piston is frictionless, both the chambers will be at same equilibrium pressure after the operation is carried out.

Considering ideality of the gas in chamber A,

`(P_(1A)V_(1A))/T_(1A)=(P_(2A)V_(2A))/T_(2A)`

`rArr`  `P_(2A) = (P_(1A)V_(1A)T_(2A))/(V_(2A)T_(1A))`

Putting the values, before and after expansion we get,

`P_(2A) = (1.0*10^5*10*S*330)/((10+x)*S*273)`

`= (10^6*330)/(273*(10+x))`

Now considering chamber B, and applying the condition of ideality for the isothermal expansion there, we get

`P_(1B)V_(1B) = P_(2B)V_(2B)`

Putting the value of `P_(2B) = P_(2A)` from above,

`1.0*10^5*10*S = (10^6*330)/((10+x)*273)*(10-x)*S`

`rArr (10-x)/(10+x)=273/330`

`rArr x=570/603`

`=0.95`

Therefore, the piston P will travel a distance of 0.95 cm towards chamber B.

 

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