Question

Calculate workdone in 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.
If process is carried out reversibly.
Compute the final volume reached by gas in this case.

Answer 1

Given that the process is reversible adiabatic.

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

$R=8.314$

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

$V_1=\frac{1\times 8.314\times 300}{1}=2494.2$ liters.

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

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

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

${\left(\frac{V_1}{V_2}\right)}^{1.66}=2\Rightarrow {\left(\frac{2494.2}{V_2}\right)}^{1.66}=2$

$1.66log\left(\frac{2494.2}{V_2}\right)=log2$

$5.638-1.66\left({logV}_2\right)=0.30$

$\frac{5.338}{1.66}=log{V}_2\Rightarrow 3.215=log{V}_2$

$V_2=1640.0lit$