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.

A cyclic process for an ideal monatomic gas (C_v = 12.5 J mol^1 K^1)is

1.	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» Solution :In the process from `1to 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)=3750joules 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