Question

An ideal gas at 300 K occupies a volume of 0.5 $m^3$ at a pressure of 2 atm. The gas expands adiabatically until its volume is 1.2 $m^3$. Next, the gas is compressed isobarically to its original volume. Finally, the pressure is increased isochorically until the gas returns to its initial state. Represent the change on the P-V diagram. Determine the temperature at the end of each transformation; and the work done during the cycle. Assume g = 1.4.

• A. 1.2
• B. 1.5
• C. 1.9
• D. 2.5

Answer 1

The correct option is C

1.9

$T={\frac{300\times {0.5}^{0.4}}{\left(12\right)}}^{0.4}orlogT=log300+0.4log0.5-0.4log1.2=2.3250orT=211.3K$

Now $\frac{P_{1}V-1}{T-1}=\frac{P_2{V}_2}{T_2}or\frac{2\times 0.5}{300}=\frac{P_2\times 1.2}{211.3}\therefore P_2=\frac{211.3}{360}=0.587atm$
Thus, pressure, temperature and volume at B are 0.587 atm, 211.3 K, and1.2$m^3.$
$ApplyingPV=\frac{m}{M}RTor\frac{PV}{T}=aconstanttoBandC\frac{0.587\times 1.2}{211.3}=\frac{0.587\times 0.5}{T}orT=88.2KThus,pressure,temperatureandvolumeatCare0.587atm,88.2K,and1.2m^{3}W-1,workdonefromAtoB=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}\Rightarrow W_1=\frac{2\times 0.5-0.587\times 12}{1.4-1}\times 1.013\times {10}^5=0.75\times {10}^{5}J=areaAB{B}^{\prime }{C}^{\prime }A{W}_{2}workdonebythegasfromBtoC=P_2(V_3-V_2)\Rightarrow W_2=0.587\times 1.013\times {10}^5\times (0.5-1.2)=-0.42\times {10}^{5}J=areaB{B}^{\prime }{C}^{\prime }CworkdonebythegasinthecycleABCA=0.75\times {10}^5-0.42\times {10}^5=0.33\times {10}^{5}J$