Question

One mole of a monoatomic ideal gas is taken through the cycle as shown in figure:

$A\to B$: adiabatic expansion

$B\to C$: cooling at constant volume

$C\to D$: adiabatic compression

$D\to A$: heating at constant volume

The pressure and temperature at A, B... etc., are$P_A,T_A,P_B,T_B$...respectively.

Given, $T_A=1000K$,

$P_B=\frac{2}{3}{P}_A$ and $P_C=\frac{1}{3}{P}_A$ calculate :

(a) The work done by the gas in the process $A\to B$ is X.

(b) The heat lost by the gas in the process $B\to C$ is Y.

(c) The temperature $T_D$ is Z.

[Given, $[\frac{2}{3}{]}^{2/5}=0.85$]
Find the value of $|X+Y+Z|$.

Answer 1

a) $W_{AB}=\frac{nR[T_A-T_B]}{\gamma -1}$

Here $n=1;R=8.32;T_A=1000K$

$\gamma =\frac{5}{3}$

To find $T_B$ we use $\frac{T_A^{\gamma }}{P_A^{\gamma -1}}=\frac{T_B^{\gamma }}{P_B^{\gamma -1}}\Rightarrow (\frac{P_A}{P_B}{)}^{\gamma -1}=(\frac{T_A}{T_B}{)}^{\gamma }=(3/2{)}^{\gamma -1}$
$T_B=850K$

$W_{AB}=\frac{1\times 8.31[1000-850]}{\frac{5}{3}-1}=1870J$

b) Heat lost : $B\to C$

$Q=n\times C_v△T=n\times C_v△(T_B-T_C)$

$T_B=850K$

To find $T_C$ we use $\frac{P_B}{T_B}=\frac{P_C}{T_C}\left(Volumeconstant\right)$

Substituting the values we get $T_C=425K$

$Q=1\times \frac{3}{2}\times 8.31\times [425-850]=-5298J$

C) Temperature $T_D$

$T_A=T_D=1000K$