A cyclic process for an ideal monatomic gas `(C_v = 12.5 J mol^1 K^1)`is represented in the figure. The temperatures at 1,2 and 3 are `300 K and 455 K`, respectively. Compute the values of `triangleQ,triangleU and triangleW` for each of the processes. The process from 2 to 3 is adiabatic.

Answer

1989.1 joules

Explanation

In the process from `1 to 2`
`triangle W = intpd V= 0` (volume remains constant)
`triangleQ = overset(T_1)underset(T_2)int(C_v)dT=C_(v)(T_(2)-T_(1))`
`triangleQ = underset(T_1)overset(T_2)intC_(v)dT = C_(v)(T_(2)-T_(1))`
12.5(600-300)=3750 joules
By the first law of thermodynamics
`triangleQ=triangleU+triangleW or triangleU=triangleQ-triangleW`
3750-0=3750 joule
In the process 2 to 3 `triangleQ=0`
(Since the process is adiabatic)
`triangleW = (R(T_2-T_3))/(gamma-1)`
`C_v = (T_2-T_3) (becauseC_v=(R)/(gamma-1))`
`=12.5(600-455)=12.5xx145=-1812.5` joules
`thereforeU=triangleQ`- `triangleW=0-1812.5=-1812.5` joules
In the process from 3 to 1, `triangleW= underset(V)overset(v_1)intpdV=p(V_1-V_3)=pV_(1)-pV_(2)`
or `triangleW=R(T_(1)-T_(3))` (becausepV=RT)
`=8.31(300-455)=-1288 joules`
`triangleQ=undersetT_(3)oversetT_(1)intC_(p)dT=C_(p)T_(1)-T_(3))=1.67xx12.5xx(300-455)`(because`gamma=(C_(p))/(C_(v))`
`=-3235.6 joules`
By the first law of thermodynamics
`triangleQ=triangleU+triangleW`
`therefore` `triangleU+triangleQ-triangleW`=(-3235.6)-(-1288)=1989.1 joules