Question

One mole of monoatomic gas is taken from point A to point B along path $A\to C\to B.$ Temperature at A is $T_0$, then heat absorbed in path ACB is

• A. $\frac{9}{2}R{T}_0$
• B. $\frac{11}{2}R{T}_0$
• C. $\frac{3}{2}R{T}_0$
• D. $\frac{15}{2}R{T}_0$

Answer 1

The correct option is B $\frac{11}{2}R{T}_0$

By, $\frac{PV}{T}=nR$----(i)
At point $A$, $\frac{P_0\times V_0}{nR}=T_A=T_0$
At point $C$, $\frac{P_0\times 2V_0}{nR}=T_C$
$\Rightarrow T_C=2T_0$
AT point $B$, $\frac{2P_0\times 2V_0}{T_B}\Rightarrow T_B=4T_0$

$\Delta Q=\Delta U+\Delta W$

Process A to C
$\Delta Q_1=n{C}_V\Delta T+P_0\Delta V$
$\Delta Q_1=1\times \frac{3}{2}R{T}_0+P_0{V}_0$
from eq. (i) $P_0{V}_0=1\times R\times T_0$
$\Delta Q_1=\frac{3}{2}R{T}_0+R{T}_0=\frac{5}{2}R{T}_0$

Process C to B
$\Delta Q_2=n{C}_V\Delta T+P_0\Delta V$
As $\Delta V=0$
so
$\Delta Q_2=n{C}_V\Delta T$
$\Delta Q_2=1\times \frac{3}{2}R\times 2T_0=3R{T}_0$

$\Delta Q=\Delta Q_1+\Delta Q_2$
$\Delta Q=\frac{5}{2}R{T}_0+3R{T}_0$
$\Delta Q=\frac{11}{2}R{T}_0$