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Question

Physics Question on Thermodynamics

One mole of a monatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature (VTV-T) diagram. The correct statement(s) is/are: [ RR is the gas constant]

A

Work done in this thermodynamic cycle (12341)\left(1\to2\to3\to4\to1\right)isW=12RT0\left|W\right|=\frac{1}{2}RT_{0}

B

The above thermodynamic cycle exhibits only isochoric and adiabatic processes

C

The ratio of heat transfer during processes 121\to2 and 23isQ12Q23=532\to3is \left|\frac{Q_{1}\rightarrow2}{Q_{2}\to3}\right|=\frac{5}{3}

D

The ratio of heat transfer during processes 121\to2 and 34isQ12Q34=123\to4is \left|\frac{Q_{1}\rightarrow2}{Q_{3}\to4}\right|=\frac{1}{2}

Answer

The ratio of heat transfer during processes 121\to2 and 23isQ12Q23=532\to3is \left|\frac{Q_{1}\rightarrow2}{Q_{2}\to3}\right|=\frac{5}{3}

Explanation

Solution

ΔQ(12)=nCPdT=n×52R×T0=5RT02\left|\Delta Q_{\left(1\rightarrow2\right)}\right|=nC_{P}dT=n\times\frac{5}{2}R\times T_{0}=\frac{5RT_{0}}{2}
ΔQ34=nCPdT=n×52RT02=5RT04\left|\Delta Q_{3\rightarrow4}\right|=nC_{P}dT=n\times\frac{5}{2}R \frac{T_{0}}{2}=\frac{5RT_{0}}{4}
ΔQ12ΔQ34=2AlsoΔQ23=32RT0\left|\frac{\Delta Q_{1\to2}}{\Delta Q_{3\rightarrow4}}\right|=2 Also\left|\Delta Q_{2\to3}\right|=\frac{3}{2}RT_{0}
(ΔQ12)(ΔQ23)=5RT0×22×3RT0=53\frac{\left(\Delta Q_{1\to2}\right)}{\left(\Delta Q_{2\rightarrow3}\right)}=\frac{5RT_{0}\times2}{2\times3RT_{0}}=\frac{5}{3}
w=ΔQw=\sum\Delta Q
=RT02=\frac{RT_{0}}{2}