Physics, asked by akashdhillonkik, 2 days ago

A chamber of the constant volume contains hydrogen gas at (0°C), the pressure of the gas is 107 kPa. What pressure will be indicated when the chamber is brought to 100°C?

Answers

Answered by Kalpesh0099

Answer:

We are doing the process at constant volume.

We are doing the process at constant volume.Applying Gay Lussac's law

We are doing the process at constant volume.Applying Gay Lussac's lawT

We are doing the process at constant volume.Applying Gay Lussac's lawT 1

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 =

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2 P

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2 P 2

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2 P 2 =

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2 P 2 = 273.15

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2 P 2 = 273.15860

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2 P 2 = 273.15860 ×373.15

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2 P 2 = 273.15860 ×373.15P

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2 P 2 = 273.15860 ×373.15P 2

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2 P 2 = 273.15860 ×373.15P 2 =1174.84torr

Please mark me as BRAINLIEST

Answered by nareshkannojia

Answer:

We are doing the process at

constant volume.

We are doing the process at

constant volume.Applying Gay

Lussac's law

We are doing the process at

constant volume.Applying Gay

Lussac's lawT

We are doing the process at

constant volume.Applying Gay

Lussac's lawT 1

We are doing the process at

constant volume.Applying Gay

Lussac's lawT 1

We are doing the process at

constant volume.Applying Gay

Lussac's lawT 1

We are doing the process at

constant volume.Applying Gay

Lussac's lawT 1P

We are doing the process at

constant volume.Applying Gay

Lussac's lawT 1 P 1

We are doing the process at

constant volume.Applying Gay

Lussac's lawT 1 P 1 We are doing the process at

constant volume.Applying Gay

Lussac's lawT 1 P 1

We are doing the process at

constant volume.Applying Gay

Lussac's lawT 1 P1 We are doing the process at

constant volume.Applying Gay

Lussac's lawT 1 P1 =

We are doing the process at

constant volume.Applying Gay Lussac's lawT 1P1 = T

We are doing the process at

constant volume.Applying Gay

Lussac's lawT 1 P1 = T2

We are doing the process at

constant volume.Applying Gay

Lussac's lawT1P1 = T2

We are doing the process at

constant volume.Applying Gay

Lussac's lawT1P1 = T2P2

We are doing the process at

constant volume.Applying Gay

Lussac's lawT 1P1 = T2P2

We are doing the process at

constant volume.Applying Gay

Lussac's lawT 1P1 = T2P2 P

We are doing the process at

constant volume.Applying Gay

Lussac's lawT 1 P1 = T2P2 P2

We are doing the process at

constant volume.Applying Gay

Lussac's lawT 1P1 = T2P2 P2

We are doing the process at

constant volume.Applying Gay

Lussac's lawT1P1 = T2P2 P2

We are doing the process at

constant volume.Applying Gay

Lussac's lawT 1P1 = T2 P2 P2= 273.15 We are doing the process at

constant volume.Applying Gay

Lussac's lawT 1 P1 = T2P2 P2=

273.15860

We are doing the process at

constant volume.Applying Gay

Lussac's lawT1P1 = T2P2 P2=

273.15860

We are doing the process at

constant volume.Applying Gay

Lussac's lawT 1 P1 = T2 P2 P2=

273.15860 x373.15

We are doing the process at

constant volume.Applying Gay

Lussac's lawT 1P1 = T2 P2 P2=

273.15860 x373.15P

We are doing the process at

constant volume.Applying Gay

Lussac's lawT1P1 = T2 P2 P2=

273.15860 x373.15P 2

We are doing the process at

constant volume.Applying Gay

Lussac's lawT1 P1 = T2 P

2 P2 = 273.15860 x373.15P 2

=1174.84torr

We are doing the process at

constant volume.Applying Gay

Lussac's lawT 1P1 = T2 P

2 P2 = 273.15860 x373.15P 2

=1174.84torr

We are doing the process at

constant volume.Applying Gay

Lussac's lawT 1 P1 = T2 P

2 P2 = 273.15860 x373.15P 2

=1174.84torr

Previous Question

Next Question

A chamber of the constant volume contains hydrogen gas at (0°C), the pressure of the gas is 107 kPa. What pressure will be indicated when the chamber is brought to 100°C?​

Answers

We are doing the process at constant volume.

We are doing the process at constant volume.Applying Gay Lussac's law

We are doing the process at constant volume.Applying Gay Lussac's lawT

We are doing the process at constant volume.Applying Gay Lussac's lawT 1

We are doing the process at constant volume.Applying Gay Lussac's lawT 1

We are doing the process at constant volume.Applying Gay Lussac's lawT 1

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 =

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2 P

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2 P 2

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2 P 2

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2 P 2 =

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2 P 2 = 273.15

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2 P 2 = 273.15860

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2 P 2 = 273.15860

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2 P 2 = 273.15860 ×373.15

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2 P 2 = 273.15860 ×373.15P

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2 P 2 = 273.15860 ×373.15P 2

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2 P 2 = 273.15860 ×373.15P 2

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2 P 2 = 273.15860 ×373.15P 2 =1174.84torr

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2 P 2 = 273.15860 ×373.15P 2 =1174.84torr

We are doing the process at constant volume.Applying Gay Lussac's lawT 1 P 1 = T 2 P 2 P 2 = 273.15860 ×373.15P 2 =1174.84torr

A chamber of the constant volume contains hydrogen gas at (0°C), the pressure of the gas is 107 kPa. What pressure will be indicated when the chamber is brought to 100°C?