One mole of a monatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature $(V - T)$ diagram. The correct statement(s) is/are: [R is the gas constant] [2019]

Question

Smart Achievers · Accepted Answer

(1, 3)

$The P - V graph of the given V - T graph is given below.$

$(1) Work done during cyclic process (1 \to 2 \to 3 \to 4 \to 1)$

$W = area enclosed in the loop = \frac{P_{0}}{2} V_{0}$

$∵$ $P_{0} V_{0} = n R T_{0}$ $∴$ $\frac{P_{0} V_{0}}{2} = \frac{n R T_{0}}{2}$

$∴$ $Work done W = \frac{n R T_{0}}{1} = \frac{R T_{0}}{2}$ $[as n = 1]$

$(2) Process 1 \to 2 is isobaric$

$Process 2 \to 3 is isochoric$

$Process 3 \to 4 is isobaric$

$Process 4 \to 1 is isochoric$

$Hence no adiabatic process is involved.$

$(3) | Δ Q_{1 \to 2} | = | n C_{p} Δ T | = | n C_{p} (2 T_{0} - T_{0}) | = | n C_{p} T_{0} |$

$| Δ Q_{2 \to 3} | = | Δ U | = | n C_{v} Δ T | = | n C_{v} T_{0} |$

$∴$ $| \frac{Δ Q_{1 \to 2}}{Δ Q_{2 \to 3}} |$ $= \frac{C_{p}}{C_{v}} = \frac{5}{3}$

$(4) | Δ Q_{3 \to 4} | = n C_{p} \frac{T_{0}}{2}$

$∴$ $| \frac{Δ Q_{1 \to 2}}{Δ Q_{3 \to 4}} |$ $= \frac{n C_{p} T_{0}}{n C_{p} \frac{T_{0}}{2}} = \frac{2}{1}$

Solution

Q.
One mole of a monatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature $(V - T)$ diagram. The correct statement(s) is/are: [R is the gas constant] [2019]

1 Work done in this thermodynamic cycle $(1 \to 2 \to 3 \to 4 \to 1)$ is $| W |$ $= \frac{1}{2} R T_{0}$

2 The above thermodynamic cycle exhibits only isochoric and adiabatic processes.

3 The ratio of heat transfer during processes $1 \to 2$ and $2 \to 3$ is $| \frac{Q_{1 \to 2}}{Q_{2 \to 3}} |$ $= \frac{5}{3}$

4 The ratio of heat transfer during processes $1 \to 2$ and $3 \to 4$ is $| \frac{Q_{1 \to 2}}{Q_{3 \to 4}} |$ $= \frac{1}{2}$

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Solution

Q. One mole of a monatomic ideal gas goes through a thermodynamic cycle, as shown in the volume versus temperature (V-T) diagram. The correct statement(s) is/are: [R is the gas constant] [2019]

1 Work done in this thermodynamic cycle (1→2→3→4→1) is |W|=12RT0