Answer to Question #214633 in Molecular Physics | Thermodynamics for Wavie

Question

A quantity of a certain perfect gas is compressed from an initial state of 0.085m3, 1 bar to a final state of 0.034m3, 3.9 bar. The specific heat at constant volume is 0.724 kJ/kg K and the specific heat at constant pressure is 1.02 kJ/kg K. The observed temperature rise is 146 K. Calculate the gas constant R, the mass of gas present, and the increase in internal energy.

Answer 1

$R=C_p-C_v=1.02-0.724=0.296 \ kJ/(kg\times K)\\ \frac{T_2}{T_1}=\frac{p_2\nu_2}{p_1\nu_1}=\frac{3.9\times 0.034}{1\times0.085}=1.56\\ Also, \ T_2-T_1=146 \ K\\ Therefore, 1.56T_1-T_1=146\\ T_1=\frac{146}{0.56}=261 \ K\\ m=\frac{p_1\nu_1}{RT_1}=\frac{1\times 10^5 \times 0.085}{0.296\times 10^3\times261}=0.14 \ kg\\ \Delta u=mc_v\Delta T=0.11\times0.724\times146=11.63 \ kJ.$