Question

One mole of an ideal monoatomic gas is taken from A to C along the path ABC. The temperature of the gas at A is $T_0$. For the process ABC:-

• A. Work done by the gas is $R{T}_0$
• B. change in internal energy of the gas is $\frac{11}{2}R{T}_0$
• C. Heat absorbed by the gas is $\frac{11}{2}R{T}_0$
• D. Heat absorbed by the gas is $\frac{13}{2}{T}_0$

Answer 1

The correct option is A Work done by the gas is $R{T}_0$

Work done by the gas=work done in $AB$ path+work done in $BC$ path.

As, $BC$ is isochoric,so, work done in $BC$ path is zero.

$\therefore$ work done by the gas=Work done in $AB$ path.

$=P_o\times ({2V}_o-V_o)$

$P_o{V}_o$ = $R{T}_o$