Question

The temperature of gas is -73°C. To what temperature should it be heated so that
a) average kinetic energy of the molecule is doubled.
b) the root mean square velocity of the molecules is doubled.​

Answer 1

Given:

The temperature of gas is -73°C.

To find:

Temperature at which:

a) average kinetic energy of the molecule is doubled.

b) the root mean square velocity of the molecules is doubled.

Calculation:

For any ideal gas , Kinetic Energy depends on Temperature as:

$\boxed{ \bf{KE \: \propto \: T }}$

So, let new temperature be T_(2)

$\rm{ \therefore \: \dfrac{(KE)_{2}}{KE} = \dfrac{T_{2}}{T} }$

$\rm{ = > \: \dfrac{2(KE)}{KE} = \dfrac{T_{2}}{( - 73 + 273)} }$

$\rm{ = > \: 2 = \dfrac{T_{2}}{( 200)} }$

$\rm{ = > \: T_{2} = 400 \: kelvin}$

$\boxed{\rm{ = > \: T_{2} = {127}^{ \circ}C }}$

Again, for any ideal gas , the RMS Velocity depends on temperature as:

$\boxed{ \bf{v \: \propto \: \sqrt{T} }}$

So , let new temperature be T_(2)

$\rm{ \therefore \dfrac{v_{2}}{v} = \dfrac{ \sqrt{T_{2}} }{ \sqrt{T} } }$

$\rm{ = > \dfrac{2v}{v} = \dfrac{ \sqrt{T_{2}} }{ \sqrt{ - 73 + 273} } }$

$\rm{ = > 2 = \dfrac{ \sqrt{T_{2}} }{ \sqrt{ - 73 + 273} } }$

$\rm{ = > 2 = \dfrac{ \sqrt{T_{2}} }{ \sqrt{ 200} } }$

$\rm{ = > 4 = \dfrac{ T_{2} }{ 200} }$

$\rm{ = > T_{2} = 800 \: kelvin}$

$\boxed{ \rm{ = > T_{2} = {527}^{ \circ} C }}$

Hope It Helps.

Answer 2

Answer:

127°C

Explanation:

the Kinetic energy gets double of the initial value