Question

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

Answer 1

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Answer 2

The temperature of the gas is$\dfrac{320}{R}^{\circ}\ C$

Explanation:

Given that,

Molecular mass = 40 g/mol

Temperature = 30°C

Velocity = 200 m/s

We need to calculate the work done on the gas

Using first law of thermodynamics

$\Delta U=\Delta Q-W_{on\ gas}$

Here, $\Delta Q=0$

$W_{on\ gas}=\dfrac{1}{2}mv^2$

Put the value into the formula

$W_{on\ gas}=\dfrac{1}{2}\times m\times(200)^2$

We need to calculate the change in internal energy

Using formula of internal energy

$\Delta U =n\times c_{v}\times\Delta T$

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

We need to calculate the rise in the temperature of the gas

Using first law of thermodynamics

$\Delta U=-W_{by\ gas}=W_{on\ gas}$

Here, $\Delta Q =0$

Put the value in the equation

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

$\dfrac{1}{M}\times\dfrac{5R}{2}\times\Delta T=\dfrac{1}{2}\times200^2$

$\dfrac{1}{40\times10^{-3}}\times\dfrac{5R}{2}\times\Delta T=2\times10^{4}$

$\Delta T=\dfrac{2\times10^{4}\times2\times40\times10^{-3}}{5R}$

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

Hence, The temperature of the gas is$\dfrac{320}{R}^{\circ}\ C$

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Topic : Rise temperature

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