
急!!!请教热力学问题!!
一摩尔单原子理想气体经历可逆循环a->b->c->a,如下:ab过程体积不变,压强增大;bc是绝热膨胀,Pb=10.0atm,Vb=1.00E-,Vc=8Vb;ca是等压...
一摩尔单原子理想气体经历可逆循环a->b->c->a,如下:ab过程体积不变,压强增大;bc是绝热膨胀,Pb=10.0atm, Vb=1.00E-, Vc=8Vb;ca是等压收缩。求(a)气体吸收的热量;(b)气体释放的热量;(c)气体做的功;(d)整个循环的效率。
英文原题: One mole of a monatomic ideal gas is taken through the reversible cycle shown below. Process bc is an adiabatic expansion, with Pb=10.0atm and Vb=1.00*10^(-3)m^3. Find (a)the energy added to the gas as heat; (b)the energy leaving the gas as heat; (c)the net work done by the gas; (d)the efficiency of the cycle.
谢谢!!! 展开
英文原题: One mole of a monatomic ideal gas is taken through the reversible cycle shown below. Process bc is an adiabatic expansion, with Pb=10.0atm and Vb=1.00*10^(-3)m^3. Find (a)the energy added to the gas as heat; (b)the energy leaving the gas as heat; (c)the net work done by the gas; (d)the efficiency of the cycle.
谢谢!!! 展开
1个回答
展开全部
bc绝热过程 p*V^γ = C
pb*Vb^(5/3) = pc*Vc^(5/3) , 求出:pc = 5/16 atm
(a)气体吸收的热量
ab过程吸热 Qab = 3/2 R (Tb - Tc) = 3/2 (pb*Vb - pc*Vc) = 1472.02 J
(b)气体释放的热量
ca过程放热 Qca = 5/2 R (Ta - Tc) = 5/2 (pa*Va - pa*Vc) = -553.984 J
(c)气体做的功 (待求)
(d)整个循环的效率
η = 1 - │Qca│/Qab = 62.3657%
(c)气体做的功
A = Qab * η = 918.036 J
pb*Vb^(5/3) = pc*Vc^(5/3) , 求出:pc = 5/16 atm
(a)气体吸收的热量
ab过程吸热 Qab = 3/2 R (Tb - Tc) = 3/2 (pb*Vb - pc*Vc) = 1472.02 J
(b)气体释放的热量
ca过程放热 Qca = 5/2 R (Ta - Tc) = 5/2 (pa*Va - pa*Vc) = -553.984 J
(c)气体做的功 (待求)
(d)整个循环的效率
η = 1 - │Qca│/Qab = 62.3657%
(c)气体做的功
A = Qab * η = 918.036 J

2025-03-07 广告
随着城市燃气管网规模的不断扩大,旧有的管理手段和管理模式都已无法满足“合理规划,科学管理,安全供气,优质服务,放心用气”的要求,力控eForceCon在燃气调度管理中心的应用,可以帮助企业提升管理。燃气调度中心SCADA系统可为管理层提供一...
点击进入详情页
本回答由力控科技提供
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询