Question

if 1 mole of oxygen at 300k expand adiabatically from 8L to 15 L. Find the temperature of the given gas

Answer 1

Answer:

TV8-1 = const T₁V₁8-²1 = T₂ V₂₂² ४.1 300 V ³² -¹ = T(8V) ³²1

300 √ 7/³ = π(8y²² 300 = 2x2T

T= 75K

2019.03.04

Answer 2

Answer:

The temperature of the given gas is 233.276K.

Explanation:

It is given that 1 mole of oxygen at 300k expand adiabatically from 8L to 15 L.

So, no of moles of oxygen= n =1 mole

no of moles of oxygen= n =1 moleV1 = 8L

no of moles of oxygen= n =1 moleV1 = 8LV1= 15L

no of moles of oxygen= n =1 moleV1 = 8LV1= 15LT1= 300K

For the Adiabatic Process, we have

$p{v}^{ \gamma } = \: constant$

$p = pressure \: of \: the \: gas \\ v = volume \: of \: the \: gas \\ \gamma = adiabatic \: constant \:$

We know that,

$pv = nRT$

$p = pressure \\ v = volume \\ n = no \: of \: moles \\ R = gas \: constant \: T = temperature$

$\: p = \frac{nRT}{v}$

putting the value of p in the Adiabatic equation, we get

$\frac{nRT}{v} \times {v}^{ \gamma } = constant$

As, value of n is 1 and R is a constant, so taking R to the R.H.S. we get,

$\frac{T}{v} \times {v}^{ \gamma } = constant$

or,

$T \times {v}^{ \gamma - 1} = constant$

or,

$T1 {v1}^{ \gamma - 1} = T2 {v2}^{ \gamma - 1}$

$we \: know \: that \: \gamma = 1 + \frac{2}{f}$

where, f is the degree of freedom.

The degree of freedom ( f ) of Oxygen = 5.

so,

$\gamma = 1 + \frac{2}{5} = \frac{7}{5}$

Then, putting the values of T1 =300K, V1= 8L, V2= 15L , we get

$300 \times {8}^{ \frac{7}{5} - 1} = T2 \times {15}^{ \frac{7}{5} - 1}$

or,

$300 \times {8}^{ \frac{2}{5} } = T2 \times {15}^{ \frac{2}{5} }$

or,

$T2 = \: \frac{300 \times 2.297}{2.954}$

or,

$T2 = 233.276K$

Conclusion: The temperature of the Oxygen gas is found to be 233.276K

For more about the Adiabatic Process following links are provided:

https://brainly.in/question/1074102?

https://brainly.in/question/7853335?