Question

The volume of a monoatomic ideal gas increases linearly with pressure as shown in the diagram. What is the increase in internal energy?

• A. $4.8\times {10}^{5}J$
• B. $3.6\times {10}^{5}J$
• C. $2.9\times {10}^{5}J$
• D. $5.6\times {10}^{5}J$

Answer 1

The correct option is A $4.8\times {10}^{5}J$

For an ideal gas:
Internal energy: $U=\frac{PV}{\gamma -1}$
Change in internal energy: $\Delta U=\frac{P_2{V}_2-P_1{V}_1}{\gamma -1}$ ....(1)
From the diagram:
$P_1=4\times {10}^5$
$V_1=0.2m^3$
$P_2=8\times {10}^5$
$V_2=0.5m^3$
For an ideal monoatomic gas, $f=3$. Thus $\gamma =\frac{C_p}{C_v}=\frac{1+\frac{f}{2}}{\frac{f}{2}}=\frac{5}{3}$
Putting the values in (1)

$\Delta U=\frac{8\times {10}^5\times 0.5-4\times {10}^5\times 0.2}{\frac{5}{3}-1}=\frac{3.2\times {10}^5}{\frac{2}{3}}=4.8\times {10}^{5}J$