Answer to Question #183564 in Mechanics | Relativity for jay

Question

An ideal gas initially having volume 2.0 x 10^-3m³ at a pressure of 1.2 x 10⁵ Pa and a temperature of 0℃ is brought around a cycle. (A) The pressure is increased to 1.5 x 10⁵ Pa without a change in volume, (B) the gas expands isothermally until it returns to its original pressure and (C) the gas is compressed at constant pressure until it returns to its original state.

What is (a) the temperature of the gas during process B, (b) the volume of gas at the end of process B and (c) the work done during the cycle.

Answer 1

Solution.

$V_1=2.0\sdot10^{-3}m^3;$

$p_1=1.2\sdot10^5Pa;$

$T_1=273K;$

$p_2=1.5\sdot10^5Pa;$

$a) V=const;$

$\dfrac{P_1}{T_1}=\dfrac{P_2}{T_2}\implies T_2=\dfrac{P_2\sdot T_1}{P_1};$

$T_2=\dfrac{1.5\sdot10^5Pa\sdot 273K}{1.2\sdot10^5Pa}=341K;$

$b) P_1V_1=P_2V_2\implies V_2=\dfrac{P_1V_1}{P_2};$

$V_2=\dfrac{1.5\sdot10^5Pa\sdot2\sdot10^{-3}m^3}{1.2\sdot10^5Pa}=2.5\sdot10^{-3}m^3;$

$c)$ $A=\nu RT_2ln\dfrac{V_2}{V_1}-p_1\Delta V;$

$\nu=\dfrac{P_1V_1}{T_1};$

$\nu=\dfrac{1.2\sdot10^5Pa\sdot 2.0\sdot10^{-3}m^3}{273K}=0.88mol;$

$A=0.88mol\sdot8.31J/(molK)441Kln\dfrac{2.5\sdot10^{-3}m^{-3}}{2\sdot10^{-3}m^{-3}}+1.2\sdot10^5\sdot0.5m^{-3}=-1635J;$

Answer: $a)T_2=341K;$

$b)$ $V_2=2.5\sdot10^{-3}m^3;$

$c)A=-1635J.$