Physics, 01.08.2019 12:00 gilliandooley1002

How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

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Answer from: kukisbae

Missing figure (and details) here:
http://i.imgur.com/l6n4Sry.jpg

Solution:
1) Let's use the ideal gas state law to find the number of moles of the gas:
p_iV_i=nRT_i

where

is the initial pressure, V_i

the initial volume of the gas, n the number of moles, R=8.31 J/mol K

and

the initial temperature of the gas.
We need to find the pressure and the volume first. The pressure is given by the spring acting on the piston, with a force $F=k \Delta x$ , where $\Delta x=5 cm=0.05 m$ is the spring compression. The area of the piston is $A=8 cm^2=8\cdot 10^{-4} m^2$ , therefore the pressure is
$p_i= \frac{F}{A}= \frac{k \Delta x}{A}= \frac{2000 N/m \cdot 0.05 m}{8\cdot 10^{-4} m^2}=1.25\cdot 10^5 Pa$
The volume is instead the length of the cylinder (L=13 cm=0.13m) multiplied by the area of the piston:
$V_i=L\cdot A=0.13 m\cdot 8\cdot 10^{-4}m^2=1.04\cdot 10^{-4}m^3$
Using these values and T_i=300 K

, we can find the number of moles:
$n= \frac{p_i V_i}{R T_i}=5.21\cdot 10^{-3} mol$

2) After the heat is added, the cylinder expands to from L=13 cm to L=16 cm. Therefore, all the conditions of the gas (p,V and T) have changed, but the number of moles is still the same. However, we can calculate the new pressure and volume of the gas.
The spring has compressed by 3 cm (because the cylinder is now 3 cm longer), so $\Delta x = 0.08 m$ . Therefore the new pressure is
$p_f= \frac{k \Delta x}{A}= \frac{2000 N/m \cdot 0.08 m}{8\cdot 10^{-4}m^2}=2\cdot 10^6 Pa$
The length of the cylinder is now L=16 cm=0.16m, therefore the new volume is
$V_f=L\cdot A=0.16 m\cdot 8\cdot 10^{-4}m^2=1.28\cdot 10^{-4}m^3$
And so, from the ideal gas law we can find the new temperature of the gas:
$T_f= \frac{p_f V_f}{n R}=591 K$

3) Now we can calculate the variation of internal energy of the gas. This is given by:
$\Delta E = n C_V \Delta T$
where $C_V= \frac{3}{2}R$ for a monoatomic gas, and $\Delta T = 591 K-300 K=291 K$ . Therefore we find
$\Delta E=5.21\cdot 10^{-3} mol \cdot ( \frac{3}{2} 8.31 J/mol K) \cdot 291 K=18.9 J$

4) However, not all the heat added is used to increase the internal energy of the gas: part of it is used to compress the spring, doing a work of
$W= \frac{1}{2}k(\Delta x_f^2-\Delta x_i^2)= \frac{1}{2}\cdot 2000 N/m ((0.08m)^2-(0.05m)^2)=3.9 J$

5) And finally, using the first thermodynamics law, we can find the total heat added to the system:
$Q=\Delta E + W=18.9J+3.9J=22.8J$

Answer from: Quest

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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