Question

Calculate the work done by adiabatic compression of one mole of an ideal gas (monoatomic) from an initial pressure of 1 atm to final pressure of 2 atm. Initial temperature =300 K.
(a) If the process is carried out reversibly
(b) If the process is carried out irreversible against 2 atm external pressure.

Compute the final volume reached by gas in two cases and describe the work graphically .

Answer 1

(a) Given that the process is reversible adiabatic.

$P_1{V}_1^r=P_2{V}_2^r(r=1.66),T_1=300K$

$R=8.314$

$P_1{V}_1=nR{T}_1$

$V_1=\frac{1\times 8.314\times 300}{1}=24.615$ liters.

$P_1{V}_1^r=P_2{V}_2^r$

$1\times V_1^r=2\times V_2^r$

${\left(\frac{V_1}{V_2}\right)}^r=2(r=1.66)$

${\left(\frac{V_1}{V_2}\right)}^{1.66}=2\Rightarrow {\left(\frac{24.615}{V_2}\right)}^{1.66}=2$

$1.66\times log\left(\frac{24.615}{V_2}\right)=log2$

$2.31-1.66\left({logV}_2\right)=0.30$

$\frac{2.09}{1.66}=log{V}_2\Rightarrow 1.259=log{V}_2$

$V_2=18.156lit$

b) For irreversible process.

$P_2{V}_2=n{RT}_2$

$T_2=\frac{2\times V_2}{1\times 0.0821}$

$V_2=\frac{T_2\times 1\times 0.0821}{2}$