One mole of an ideal gas monatomic is taken from A to C along path ABC- The temperature of gas is T 0 at A- then for process ABCA- Work done by gas is RT0-B- Change in internal energy is 112 RT 0-C- Heat absorb by gas is 112 RT 0-D- Heat absorb by gas is 132 RT 0 -

The correct options are A Work done by gas is nbsp;RT0. C Heat absorb by gas is nbsp;112RT0.Area under PV graph= work done=P0V0=RT0 P_0V_0/T_0=2P_0.2V_0/T_ ...

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Standard XII

Physics

Specific Heat Capacity

One mole of a...

Question

One mole of an ideal gas (monatomic) is taken from A to C along path ABC. The temperature of gas is $T_{0}$ at A, then for process ABC

Work done by gas is

R T_{0} .

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Change in internal energy is

\frac{11}{2} R T_{0} .

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Heat absorb by gas is

\frac{11}{2} R T_{0} .

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Heat absorb by gas is

\frac{13}{2} R T_{0} .

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Solution

The correct options are
A Work done by gas is $R T_{0} .$
C Heat absorb by gas is $\frac{11}{2} R T_{0} .$
Area under PV graph= work done= $P_{0} V_{0} = R T_{0}$

$\frac{P_{0} V_{0}}{T_{0}} = \frac{2 P_{0} .2 V_{0}}{T_{C}} ⟹ T_{C} = 4 T_{0}$

Since, $Δ U = n C_{V} Δ T$
$Δ U = 1 \times \frac{3}{2} R (4 T_{0} - T_{0}) Δ U = \frac{9}{2} R T_{0} Q = Δ U + W By putting values Q = \frac{11}{2} R T_{0}$

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One mole of an ideal gas (monatomic) is taken from A to C along path ABC. The temperature of gas is T0 at A, then for process ABC

The correct options are A Work done by gas is RT0. C Heat absorb by gas is 112RT0.Area under PV graph= work done=P0V0=RT0 P0V0T0=2P0.2V0TC⟹TC=4T0 Since, ΔU=nCVΔT ΔU=1×32R(4T0−T0)ΔU=92RT0Q=ΔU+WBy putting valuesQ=112RT0

One mole of an ideal gas (monatomic) is taken from A to C along path ABC. The temperature of gas is $T_{0}$ at A, then for process ABC