a diatomic gas of molecular mass 40 gram per mole is filled in a rigid container at temperature 30 degree Celsius it is moving with velocity 200 metre per second if it is suddenly stopped the rise in temperature of gas is
Answers
The rise in temperature of gas is 320 / R° C
Explanation:
=> It is given that,
Velocity of gas, v = 200 m/s
Molecular mass of gas, M = 40g/mole = 40 * 10⁻³ kg/mole
Mass of gas enclosed = m
Rise in temperature of gas, ΔT = ?
=> Now, According to the first law of thermodynamics for the gas,
ΔU = ΔQ - W(by gas)
As there is no exchange of heat energy here, thus ΔQ = 0
ΔU = - W(by gas) ...(1)
But, ΔU = n * c(v) * ΔT
ΔU = m/M * 5R/2 * ΔT ...(2)
W(on gas) = 1/2 mv²
W(on gas) = 1/2 m(200)² ...(3)
[As ΔU = - W(by gas) = W(on gas)]
=> Putting the value of eq(2) and (3) into eq (1), we get
m/M * 5R/2 * ΔT = 1/2 m(200)²
1/M * 5R/2 * ΔT = 1/2 (200)²
1/M * 5R/2 * ΔT = 2 * 10⁴
1 / 40 * 10⁻³ * 5R/2 * ΔT = 2 * 10⁴
ΔT = 2 * 10⁴ * 40 * 10⁻³ * 2 / 5R
ΔT = 1600 / 5R
ΔT = 320 / R° C
Thus, the rise in temperature of gas is 320 / R° C
Learn more:
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