Question

A diatomic gas of molecular mass $40g/mol$ is filled in a rigid container at temperature ${30}^{o}C$. It is moving with velocity $200m/s$. If it is suddenly stopped, the rise in the temperature of the gas is

• A. $\frac{32}{R}{}^{o}C$
• B. $\frac{320}{R}{}^{o}C$
• C. $\frac{3200}{R}{}^{o}C$
• D. $\frac{3.2}{R}{}^{o}C$

Answer 1

The correct option is B $\frac{320}{R}{}^{o}C$

$\text{Given:-}$ Molecular mass $=40g/mol$

Temperature $={30}^{\circ }C$

Velocity $=200m/s$

$\text{Solution:-}$

Using first law of thermodynamics

$\Delta U=\Delta Q-W_{ongas}$

Here, $\Delta Q=0$

$W_{ongas}=\frac{1}{2}m{v}^2$

Put the value into the formula

$W_{ongas}=\frac{1}{2}\times m\times {200}^2$

Using formula of internal energy

$\Delta U=n\times C_V\times \Delta T$

$\Delta U=\frac{m}{M}\times \frac{5R}{2}\times \Delta T$

Using first law of thermodynamics

$\Delta U=-W_{bygas}=W_{ongas}$

Here, $\Delta Q=0$

Put the value in the equation

$\frac{m}{M}\times \frac{5R}{2}\times \Delta T=\frac{1}{2}\times m\times {200}^2$

$\frac{1}{M}\times \frac{5R}{2}\times \Delta T=\frac{1}{2}\times {200}^2$

$\Rightarrow \Delta T=\frac{2\times {10}^4\times 2\times 40\times {10}^{-3}}{5R}$

$\Rightarrow \Delta T={\frac{320}{R}}^{\circ }C$

$\text{Hence the correct option is B}$