m grams of a gas of molecular weight M is flowing in an isolated tube velocity v. If the gas flow is suddenly stopped the rise in the temperature is: (y = ratio of specific heats, R = universal gas constant, J = mechanical equivalent of heat).

Question

m grams of a gas of molecular weight M is flowing in an isolated tube velocity v. If the gas flow is suddenly stopped the rise in the temperature is: (y = ratio of specific heats, R = universal gas constant, J = mechanical equivalent of heat). (A)  (Mv2(&gamma; -1)/2RJ)  (B)  (m/M)(v2(&gamma; -1)/2RJ)  (C)  (mv2&gamma; /2RJ)  (D)  (Mv2&gamma; /2RJ)

Tardigrade Admin · Accepted Answer

Δ U=(μ RJΔ Tau /(&gamma; -1)) ⇒ Δ T=((&gamma; -1)Δ U/μ RJ) ⇒ Δ T=((&gamma; -1)× (1/2)mv2/(m/M)RJ) =(M(&gamma; -1)v2/2RJ)=(Mv2(&gamma; -1)/2RJ)

Q. m grams of a gas of molecular weight M is flowing in an isolated tube velocity v. If the gas flow is suddenly stopped the rise in the temperature is: (y = ratio of specific heats, R = universal gas constant, J = mechanical equivalent of heat).

$\frac{M v ^{2} ( γ - 1 )}{2 R J}$

$\frac{m}{M} \frac{v ^{2} ( γ - 1 )}{2 R J}$

$\frac{m v ^{2} γ}{2 R J}$

$\frac{M v ^{2} γ}{2 R J}$

$Δ U = \frac{μ R J Δ T}{( γ - 1 )}$ $\Rightarrow$ $Δ T = \frac{( γ - 1 ) Δ U}{μ R J}$ $\Rightarrow$ $Δ T = \frac{( γ - 1 ) \times \frac{1}{2} m v ^{2}}{( m / M ) R J}$ $= \frac{M ( γ - 1 ) v ^{2}}{2 R J} = \frac{M v ^{2} ( γ - 1 )}{2 R J}$

Q. m grams of a gas of molecular weight M is flowing in an isolated tube velocity v. If the gas flow is suddenly stopped the rise in the temperature is: (y = ratio of specific heats, R = universal gas constant, J = mechanical equivalent of heat).

2RJMv2(γ−1)​

Mm​2RJv2(γ−1)​

2RJmv2γ​

2RJMv2γ​

ΔU=(γ−1)μRJΔT​ ⇒ ΔT=μRJ(γ−1)ΔU​ ⇒ ΔT=(m/M)RJ(γ−1)×21​mv2​ =2RJM(γ−1)v2​=2RJMv2(γ−1)​

$\frac{M v ^{2} ( γ - 1 )}{2 R J}$

$\frac{m}{M} \frac{v ^{2} ( γ - 1 )}{2 R J}$

$\frac{m v ^{2} γ}{2 R J}$

$\frac{M v ^{2} γ}{2 R J}$

$Δ U = \frac{μ R J Δ T}{( γ - 1 )}$ $\Rightarrow$ $Δ T = \frac{( γ - 1 ) Δ U}{μ R J}$ $\Rightarrow$ $Δ T = \frac{( γ - 1 ) \times \frac{1}{2} m v ^{2}}{( m / M ) R J}$ $= \frac{M ( γ - 1 ) v ^{2}}{2 R J} = \frac{M v ^{2} ( γ - 1 )}{2 R J}$