- #1
vitaly
- 48
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Hi, I need some help with my thermal physics. I posted these questions on the high school forum (I'm just a junior in high school), but my physics teacher takes questions out of college-level books, so I thought it would be more appropriate if I posted here.
The two questions are:
1. An aluminum cube 20 cm on a side is heated from 50° C to 150° C in a chamber at atmospheric pressure. Determine the work done by the cube and the change in its internal energy. If the same process was carried out in a vacuum, what would be the change in internal energy?
and
2. Determine the amount of work needed to compress 4.0 g of oxygen at STP down to 1/3 its original volume, keeping the temperature constant. Assume it behaves as an ideal gas.
I gave the it a shot and attempted to solve the questions. My work is shown below. The problem is I have no idea whether or not it is correct. If anybody could guide me through and tell me where I went wrong (and why), that would be very helpful. Thank you.
My work:
W = PDV
1 atm = 1.01x10^5 Pa (process is isobaric; pressure is constant.)
Volume = 20 cm^3 = 20 cm^3 x (1 m^3/100 cm^3) = 8x10^-3 m^3
W = (1.01x10^5 Pa)(8x10^-3 m^3)
= 808 J
Work is 808 Joules
DU = Q - W
Q = cmDT
Specific heat capacity (c) of aluminum is 900 J/(kg)(K)
Mass (m) is 0.0269 kg (26.9 g/1000)
150 degrees - 50 degrees = 100 degrees = 373 K
Q = (900 J/(kg)(K))(.0269kg)(373 K)
= 9030 J
DU = Q - W
= 9030 J - 808 J
= 8222 J
Change in internal pressure is 8222 J
If the process was carried out in a vacuum, the work (w) would be zero. Thus, the internal energy would be the Q - 0, which is equal to 9030 J.
2. STP constants: V = 22.4 L and P = 1.01x10^5 Pa
W = nRTV(f)/V(I) or PV(V(f))/(V(I))
N = number of moles = 4 g * (1 mole/16 g) = .25 mole
Volume = .25 mole * (22.4 L/1 mole) = 5.6 L
W = PV(F)/(I)
= (1.01x10^5 Pa)(5.6 L)(1)/3
= 1.8x10^5 J
The two questions are:
1. An aluminum cube 20 cm on a side is heated from 50° C to 150° C in a chamber at atmospheric pressure. Determine the work done by the cube and the change in its internal energy. If the same process was carried out in a vacuum, what would be the change in internal energy?
and
2. Determine the amount of work needed to compress 4.0 g of oxygen at STP down to 1/3 its original volume, keeping the temperature constant. Assume it behaves as an ideal gas.
I gave the it a shot and attempted to solve the questions. My work is shown below. The problem is I have no idea whether or not it is correct. If anybody could guide me through and tell me where I went wrong (and why), that would be very helpful. Thank you.
My work:
W = PDV
1 atm = 1.01x10^5 Pa (process is isobaric; pressure is constant.)
Volume = 20 cm^3 = 20 cm^3 x (1 m^3/100 cm^3) = 8x10^-3 m^3
W = (1.01x10^5 Pa)(8x10^-3 m^3)
= 808 J
Work is 808 Joules
DU = Q - W
Q = cmDT
Specific heat capacity (c) of aluminum is 900 J/(kg)(K)
Mass (m) is 0.0269 kg (26.9 g/1000)
150 degrees - 50 degrees = 100 degrees = 373 K
Q = (900 J/(kg)(K))(.0269kg)(373 K)
= 9030 J
DU = Q - W
= 9030 J - 808 J
= 8222 J
Change in internal pressure is 8222 J
If the process was carried out in a vacuum, the work (w) would be zero. Thus, the internal energy would be the Q - 0, which is equal to 9030 J.
2. STP constants: V = 22.4 L and P = 1.01x10^5 Pa
W = nRTV(f)/V(I) or PV(V(f))/(V(I))
N = number of moles = 4 g * (1 mole/16 g) = .25 mole
Volume = .25 mole * (22.4 L/1 mole) = 5.6 L
W = PV(F)/(I)
= (1.01x10^5 Pa)(5.6 L)(1)/3
= 1.8x10^5 J