Question

$2$ moles of an ideal monatomic gas is carried from a state $(P_0, V_0)$ to a state $(2P_0, 2V_0)$ along a straight line path in a P-V diagram. The amount of heat absorbed by the gas in the process is given by

• A. $3P_{0} V_{0}$
• B. $\frac{9}{2}P_{0} V_{0}$
• C. $6P_{0} V_{0}$
• D. $\frac{3}{2}P_{0} V_{0}$

Answer 1

Answer: (C)

The internal energy
$\Delta U=n C_{v} \Delta T$
$C_{v}=$ Specific heat of gas at constant volume
$\Rightarrow \Delta U =n \cdot \frac{3 R}{2}\left(\frac{4 p_{0} V_{0}}{n R}-\frac{p_{0} V_{0}}{n R}\right) $
$=n \cdot \frac{3 R}{2}\left(\frac{4 p_{0} V_{0}-p_{0} V_{0}}{n R}\right) $
$=n \cdot \frac{3 R}{2} \cdot \frac{3 p_{0} V_{0}}{n R} $
$=\frac{9}{2} p_{0} V_{0} \ldots$(i)
Work done by the gas
$W=\left(2 p_{0}+p_{0}\right) \frac{V_{0}}{2}=\frac{3 p_{0} V_{0}}{2}\ldots$(ii)
From first law of thermodynamics,
$\Delta Q =d W+d U $
$=\frac{3 p_{0} V_{0}}{2}+\frac{9}{2} p_{0} V_{0}$
[from Eqs. (i) and (ii)]
$=\frac{3 p_{0} V_{0}}{2}+\frac{9}{2} p_{0} V_{0} $
$=\frac{12 p_{0} V_{0}}{2}=6 p_{0} V_{0}$