Question

A cylinder is closed by a piston connected to a spring of constant 1.60 103 N/m. With the spring relaxed, the cylinder is filled with 5.00 L of gas at a pressure of 1.00 atm and a temperature of 20.0

Answer 1

The height is 0.247 m.

Step-by-step explanation:

Given that,

Spring constant = 1.60\times10^3\ N/m[/tex]

Pressure = 1.00 atm

Temperature = 20.0°C

Suppose if the piston has a cross section area of 0.0120 m² and negligible mass, how high will it rise when the temperature is raised to 250°C

The equation of the pressure is

`P'=P_(0)+(kh)/(A)`...(I)

The equation of the volume is

`V'=V_(0)+Ah`....(II)

We need to calculate the height

Using equation of ideal gas

`(P_(0)V)/(T)=(P'V')/(T')`

`P'V'=P_(0)V(T')/(T)`

Put the value of P' and V' from equation (I) and (II)

`P_(0)+(kh)/(A)* V_(0)+Ah=P_(0)V(T')/(T)`

Put the value into the formula

`1.013*10^(5)+(1.60*10^(3)* h)/(0.0120 )*5.00*10^(-3)+0.0120 * h=1.013*10^(5)*5.00*10^(-3)*(523)/(293)`

`1.013*10^(5)*5.00*10^(-3)+1.013*10^(5)*0.0120* h+1600h^2=904.09`

`1600h^2+1215.6h=904.09-506.5`

`1600h^2+1215.6h=397.59`

`h=0.247\ m`

Hence, The height is 0.247 m.