A 6.0-cm-diameter cylinder of nitrogen gas has a 4.0-cm-thick movable copper pis

Question

A 6.0-cm-diameter cylinder of nitrogen gas has a 4.0-cm-thick movable copper piston. The cylinder is oriented vertically, as shown in the figure, and the air above the piston is evacuated. When the gas temperature is 20?C, the piston floats 20cm above the bottom of the cylinder.

a) What is the gas pressure (Pa)?

A 6.0-cm-diameter cylinder of nitrogen gas has a 4.0-cm-thick movable copper piston. The cylinder is oriented vertically, as shown in the figure, and the air above the piston is evacuated. When the gas temperature is 20?C, the piston floats 20cm above the bottom of the cylinder. What is the gas pressure (Pa)? How many gas molecules are in the cylinder? Then 2.0J of heat energy are transferred to the gas. What is the new equilibrium temperature of the gas (*C)? What is the final height of the piston (cm)? How much work is done on the gas as the piston rises (J)?

Explanation / Answer

a)
pressure = mg / A = density * volume * g = density * h * A * g / A
= density *height * g
= 8960 * 0.04 * 9.8
= 3512 Pascals
------------------------------------------------
b) number of molecules can be found using theideal gas law p V = n R T
n = p V / RT
V = pi r2 H = pi * 0.03^2* 0.20 = 0.0005655
n = 3512 * 0.0005655 / 8.31 * 293
= 0.0008157 moles
-------------------------------------------------
c)for a diatomic gas, with heat added atconstant pressure
heat added = (7/2) n R delta T
2.0 = 3.5 *0.0008157 * 8.31 * deltaT
deltaT = 84.30 C
new equilibrium temp = 20 +84.30 = 104.30 degC
-----------------------------------------------
d) work done = p delta V and work done = n R deltaT so
p deltaV =n R deltaT
p A deltah = n R deltaT
A = pi r^2 = 0.002827 m2
3512 * 0.002827 * delta h = 0.0008157 * 8.31 * 84.30
deltah = 0.05755 meters = 5.76cm
-------------------------------------------------
e) work done on piston = m g delta h
= volume * density * g * deltah
= 0.04 * pi * 0.03^2 * 8960 * 9.8 *0.05755
= 0.571 J
d) work done = p delta V and work done = n R deltaT so
p deltaV =n R deltaT
p A deltah = n R deltaT
A = pi r^2 = 0.002827 m2
3512 * 0.002827 * delta h = 0.0008157 * 8.31 * 84.30
deltah = 0.05755 meters = 5.76cm
-------------------------------------------------
e) work done on piston = m g delta h
= volume * density * g * deltah
= 0.04 * pi * 0.03^2 * 8960 * 9.8 *0.05755
= 0.571 J
3512 * 0.002827 * delta h = 0.0008157 * 8.31 * 84.30
deltah = 0.05755 meters = 5.76cm
-------------------------------------------------
e) work done on piston = m g delta h
= volume * density * g * deltah
= 0.04 * pi * 0.03^2 * 8960 * 9.8 *0.05755
= 0.571 J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.