Question

急！！！！翻译高手请进！！！！！另有悬赏！！！！

Since the question concerns the entire apparatus, the system is taken as the gas, piston, and cylinder. No work is done during the process, because no force external to the system moves, and no heat is transferred through the vacuum surrounding the apparatus. (101.3)(0.2-0.1)kPaHence Q and W are zero, and the total energy the system does not change, Without further information we can say nothing about the distribution of energy among the parts of system. This may well be different than the initial distribution.
Example2.3
If the process described in Ex. 2.2 is repeated, not in a vacuum but in air at atmospheric pressure of 101.3kPa, what is the energy change of the apparatus? Assume the rate of heat exchange between the apparatus and the surrounding air is slow compared with the rate at which the process occurs.
Solution 2.3
The system is chosen exactly as before, but in this case work is done by the system in pushing back the atmosphere. This work is given by the product of the force exerted by atmospheric pressure on the back side of the piston and the displacement of the piston. If the area of piston is A, the force is F=PatmA. The displacement of the piston is equal to the volume change of the gas divided by the area of the piston, or△l=△Vt/A. Work is done by the system on the surroundings. By Eq.(1.1),

Work done by system=F △l=Patm△Vt kN
Since W is work done on the system, it is the negative of this result:
W=-10.13kN m=-10.13kJ
Heat transfer between the system and surrounding is also possible in this case, but the problem is worked for the instant after the process has occurred and before appreciable heat transfer has had time to take place. Thus Q is assumed to be zero in Eq. (2.2), giving △(Energy of the system)=Q + W=0-10.13=-10.13 kJ
The total energy of the system has decreased by an amount equal to work done on the surroundings.

Answer 1

因为问题有关整个用具 系统被采取作为气体活塞 ,并且圆筒 在过程期间，工作没有被完成 因为对系统的没有力量外部 移动 ,并且热没有通过围拢用具的真空转移 ,(101.3) (0.2-0.1) kPaHence Q和W零，并且系统不改变的总能， 没有详细信息我们什么都不可以说关于能量的发行在系统之中的部分的。 这跟最初的发行说不定不同。
Example2.3
在前描述的过程。 2.2被重复什么，不在真空，而且在空气上以大气压101.3kPa，是能量变化用具？ 假设热交换的率在用具和周围的空气之间的是慢的比较过程发生的率。
Solution 2.3

系统正确地被选择前面，但是在这种情况下工作由在推回大气的系统完成。 这工作由在活塞的后部和活塞的位移的大气压施加的力量的产品给。 如果活塞区域是A，力量是F=PatmA。 活塞的位移与活塞， or△l=△Vt/A.工作的区域划分的气体的容积变化是相等的由在周围的系统完成。 Eq。(1.1).

system=F △l=Patm△Vt kN完成的工作
Since W是在系统完成的工作，它是这个结果阴性：
W=-10.13kN m=-10.13kJ 在系统和围拢之间的Heat调动在这种情况下也是可能的，但是问题工作在瞬时，在过程发生了之后，并且，在看得出的热传递有时间发生之前。 因而Q假设是调整归零Eq。 (2.2)，给△ (系统的能量) =Q + W=0-10.13=-10.13 kJ
在系统和围拢之间的热传递在这种情况下也是可能的，但是问题工作在瞬时，在过程发生了之后，并且，在看得出的热传递有时间发生之前。 因而Q假设是调整归零Eq。 (2.2)，给△ (系统的能量) =Q + W=0-10.13=-10.13 kJ 系统的The总能减少了由数量相等与在周围完成的工作。