Two moles of an ideal monoatomic gas undergoes a process A to B as shown in the V-T diagram. Choose the correct statement(s). [R is gas constant]

For the given process,

P V^{2}

is constant.

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For the given process,

\frac{P_{A} V_{A}}{T_{A}} = \frac{P_{B} V_{B}}{T_{B}}

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For the process,

P_{A} V_{A} = P_{B} V_{B}

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Molar heat capacity for gas is

\frac{R}{2}

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Solution

The correct options are
B For the given process, $P V^{2}$ is constant.
C For the given process, $\frac{P_{A} V_{A}}{T_{A}} = \frac{P_{B} V_{B}}{T_{B}}$ .
D Molar heat capacity for gas is $\frac{R}{2}$ .
Two moles of an ideal monoatomic gas undergoes a process A to B as shown in the V-T diagram.
The correct statements are given by options (A) and (B) .
(A) For the given process, $P V^{2}$ is constant. Since $V T =$ constant
$P V^{2} = P V \times V$
$P V = n R T$
$P V^{2} = n R T \times V = V T \times n R =$ constant.
(B) For the given process, $\frac{P_{A} V_{A}}{T_{A}} = \frac{P_{B} V_{B}}{T_{B}}$ .
This is true for an ideal gas as $P V = n R T$ so
$\frac{P V}{T} = n R =$ constant.
(D) Molar heat capacity at constant pressure for monoatomic gas is $\frac{5 R}{2}$
Molar heat capacity at constant volume for monoatomic gas is $\frac{3 R}{2}$

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