One mole of an ideal gas monoatomic is taken from A to C along the path ABC- The temperature of the gas is T 0 at A- Then for the process ABCA- Work done by gas is RT 0B- Change in internal energy is 11-2 RT 0C- Heat absorbed by gas is 11-2 RT 0D- Heat absorbed by gas is 13-2 RT 0

The correct options are A Work done by gas is RT_o C Heat absorbed by gas is nbsp;11/2RT_oWork done by gas for AB = Area under the line AB = P_o(2V_o V_o)=P_o ...

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Standard XII

Physics

Ideal Gas Equation

One mole of a...

Question

One mole of an ideal gas (monoatomic) is taken from A to C along the path ABC. The temperature of the gas is $T_{o}$ at A. Then for the process ABC

Work done by gas is

R T_{o}

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Change in internal energy is

\frac{11}{2} R T_{o}

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Heat absorbed by gas is

\frac{11}{2} R T_{o}

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Heat absorbed by gas is

\frac{13}{2} R T_{o}

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Solution

The correct options are
A Work done by gas is $R T_{o}$
C Heat absorbed by gas is $\frac{11}{2} R T_{o}$
Work done by gas for AB = Area under the line AB = $P_{o} (2 V_{o} - V_{o}) = P_{o} V_{o}$

For 1 mole of an ideal gas, $P_{o} V_{o} = R T_{o}$

Work done by gas for process BC = 0 ( $∵$ Volume is constant)

Therefore, work done by gas for ABC = Work done by gas for AB + Work done by gas for BC
Thus, total work done = $R T_{o}$

Internal energy change for ABC can be calculated by adding the internal energy change for AB and BC or directly between A and C because internal energy is point function which depends on final and initial state.

Thus, internal energy change for AC,
$Δ U = n C_{v} d t = n C_{v} (T_{C} - T_{A})$
For an ideal monoatomic gas,
$C_{v} = \frac{3}{2} R$

Now AB is a constant pressure process, thus
$V \propto T$
$∴ \frac{V_{B}}{V_{A}} = \frac{T_{B}}{T_{A}}$
Now $V_{B} = 2 V_{o}, V_{A} = V_{o} and T_{A} = T_{o}$
$\Rightarrow T_{B} = 2 T_{o}$

Now BC is a constant volume process, thus
$P \propto T$
$∴ \frac{P_{C}}{P_{B}} = \frac{T_{C}}{T_{B}}$
Now $P_{B} = P_{o}, P_{C} = 2 P_{o} a n d T_{B} = 2 T_{o}$
$\Rightarrow T_{C} = 4 T_{o}$

Thus, internal energy for ABC,
$Δ U = n C_{v} (T_{C} - T_{A})$
$Δ U = 1 \times \frac{3}{2} R (4 T_{o} - T_{o}) = \frac{9}{2} R T_{o}$
Since $Q = U + W$
For ABC, $Q = \frac{9}{2} R T_{o} + R T_{o} = \frac{11}{2} R T_{o}$

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One mole of an ideal gas (monoatomic) is taken from A to C along the path ABC. The temperature of the gas is To at A. Then for the process ABC

One mole of an ideal gas (monoatomic) is taken from A to C along the path ABC. The temperature of the gas is $T_{o}$ at A. Then for the process ABC