A vessel containing one mole of a monatomic ideal gas (molecular weight = 20 g mol^{–1}) is moving on a floor at a speed of 50 ms^{–1}. The vessel is stopped suddenly. Assuming that the mechanical energy lost has gone into the internal energy of the gas, find the rise in its temperature.

**Given:**

number of moles, n = 1

Specific heat at constant temperature, C_{v}(monoatomic gas) = 3R/2 = 12.471 J/mol/K

R = universal gas constant = 8.314 J/mol/K

Initial velocity(v_{i}) = 50 m/s

Final velocity(v_{f}) = 0

Molecular weight(m) = 20 g/mol = 0.02 kg/mol

**Formula Used:**

i. Change in internal energy(dU) = nC_{v}dT,

Where,

n is the number of the moles of the gas,

C_{v} is the heat capacity at the constant volume

dT = rise in temperature.

ii. Mechanical energy lost,

Where,

m is the molecular weight of the gas in kg

v_{i} is the initial velocity,

v_{f} is the final velocity,

equating equation (i) and (ii), we get

Putting the values in the above equation, we get

=> dT = rise in temperature ~ 2 K (Answer).

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