Question

About $0.014kg$ of nitrogen is enclosed in a vessel temperature of ${27}^{\circ }C$. How much heat has to be transferred to the gas to double the rms speed of its molecules? $(R=2cal/molK)$.

• A. ${\left(\Delta Q\right)}_v=2250cal$
• B. ${\left(\Delta Q\right)}_v=1225cal$
• C. ${\left(\Delta Q\right)}_v=4500cal$
• D. ${\left(\Delta Q\right)}_v=9000cal$

Answer 1

The correct option is A ${\left(\Delta Q\right)}_v=2250cal$

As gas is enclosed in the cylinder, $V=constant$
${\left(\Delta Q\right)}_v=\mu C_v\Delta T$
Here, $\mu =\frac{0.014\times {10}^3}{28}=\frac{1}{2}mol$
And as nitrogen is diatomic, $C_v=\frac{5}{2}R$,
Further, as according to the given problem,
$\frac{{\left(v_{rms}\right)}_2}{{\left(v_{rms}\right)}_1}=\sqrt{\frac{T_2}{T_1}}=2,$ i.e., $T_2=4T_1$
$\Delta T=4T_1-T_1=3T_1=3\times 300=900K$
${\left(\Delta Q\right)}_v=\frac{1}{2}\times \frac{5}{2}\times 2\times 900=2250cal$