One mole of an ideal gas with $\frac{C_{p}}{C_{v}} = γ$ at absolute temperature $T_{1}$ is adiabatically compressed from an initial pressure
$P_{1}$ to a final pressure $P_{2}$ . The resulting temperature $T_{2}$ of the gas is given by

$T_{2} = T_{1} {(\frac{P_{2}}{P_{1}})}^{\frac{γ}{γ - 1}}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$T_{2} = T_{1} {(\frac{P_{2}}{P_{1}})}^{\frac{γ - 1}{γ}}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$T_{2} = T_{1} {(\frac{P_{2}}{P_{1}})}^{γ}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$T_{2} = T_{1} {(\frac{P_{2}}{P_{1}})}^{γ - 1}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
$T_{2} = T_{1} {(\frac{P_{2}}{P_{1}})}^{\frac{γ - 1}{γ}}$

$F o r a n a d i a b a t i c p r o c e s s, P V^{γ} = c o n s t a n t . T h e i d e a l g a s e q u a t i o n i s P V = R T o r V = \frac{R T}{P} . T h e r e f o r e, w e h a v e P {(\frac{R T}{P})}^{γ} = c o n s t a n t o r \frac{T^{γ}}{P^{γ - 1}} = c o n s t a n t (∵ R i s a c o n s t a n t) ∴ \frac{T_{2}^{γ}}{P_{2}^{γ - 1}} = \frac{T_{1}^{γ}}{P_{1}^{γ - 1}} o r T_{2} = T_{1} {(\frac{P_{2}}{P_{1}})}^{\frac{γ - 1}{γ}}$

Suggest Corrections

Similar questions

One mole of an ideal gas with $\frac{C_{p}}{C_{v}} = γ$ at absolute temperature $T_{1}$ is adiabatically compressed from an initial pressure
$P_{1}$ to a final pressure $P_{2}$ . The resulting temperature $T_{2}$ of the gas is given by

Q. A gas with

\frac{C_{P}}{C_{V}} = γ

goes from an initial state

(p_{1}, V_{1}, T_{1})

to a final state

(p_{2}, V_{2}, T_{2})

through an adiabatic process. The work done by the gas is

Q. An ideal gas at pressure P is adiabatically compressed so that its density becomes n times the initial value. The final pressure of the gas will be

(γ = \frac{C_{p}}{C_{v}})

Q. A given mass of a gas at

0^{o} C

is compressed reversibly and adiabatically to a pressure

20

times the initial value. Calculate the final temperature of the gas.

[\frac{C_{P}}{C_{V}} = 1.42]

Pressure-temperature relationship for an ideal gas undergoing adiabatic change is $γ = \frac{C_{p}}{C_{v}}$

Join BYJU'S Learning Program

One mole of an ideal gas with CpCv=γ at absolute temperature T1 is adiabatically compressed from an initial pressure P1 to a final pressure P2. The resulting temperature T2 of the gas is given by

The correct option is B T2=T1(P2P1)γ−1γ For an adiabatic process, PVγ=constant. The ideal gas equation is PV=RT or V=RTP. Therefore, we haveP(RTP)γ=constantor TγPγ−1=constant (∵R is a constant)∴Tγ2Pγ−12=Tγ1Pγ−11or T2=T1(P2P1)γ−1γ

One mole of an ideal gas with $\frac{C_{p}}{C_{v}} = γ$ at absolute temperature $T_{1}$ is adiabatically compressed from an initial pressure
$P_{1}$ to a final pressure $P_{2}$ . The resulting temperature $T_{2}$ of the gas is given by