Thermal Properties of Matter Test - 55

Convection is the process of heat transfer in fluids by the actual motion of matter. Hence, a medium is required for convection process. As vacuum is created inside a bulb , hence heat transfer to glass bulb from filament is through radiation. 

Hint: Using ideal gas equation rearrange the given equation and then differentiate.

Where P  - Pressure, T - Temperature

Consider the ideal gas equation and substitute the value of pressure in the given equation.

Explanation of Correct Option:

Step 1: Consider the formulas mentioned

Expression of is given by,

Volume expansion can be expressed as:

 $$V={{V}_{0}}(1+\gamma t)$$

 Where V is the volume at temperature t

$${{V}_{0}}$$ Is the volume at temperature t = 0

$$\gamma $$ is the coefficient of volume expansion and t is temperature.

 

Step 2: Consider the ideal gas expansion

Ideal gas is expansion is given as

 $$P{{T}^{2}}=\text{constant}$$….. (1)

 

Step 3: Consider the ideal gas equation,

$$PV=nRT $$

$$\dfrac{PV}{T}=\text{constant}$$

$$P=\dfrac{\text{constant}\times T}{V}......\left( 2 \right)$$

Coefficient of volume expansion can be calculated by the values of gas denoted by $$\gamma$$

 

Step 4: Coefficient of volume expansion of gas is given by

$$V={{V}_{0}}(1+\gamma t).......\left( 3 \right)$$

Substitute the  equation (2) in equation (1),

$$\dfrac{\text{constant}\times T}{V}\times {{T}^{2}}=\text{constant}$$

 $$ \dfrac{{{T}^{3}}}{V}=\text{constant=k}$$

$$ V=\dfrac{1}{k}{{T}^{3}}={{k}^{1}}{{T}^{3}}.....\left( 4 \right)\text{     }\left( \because {{k}^{1}}=\dfrac{1}{k} \right)$$

 

Step 5: Differentiate equation (3) with respect to t

On differentiating with respect to t

$$ \dfrac{dV}{dt}={{V}_{0}}(\gamma .1) $$

$$\dfrac{dV}{dt}={{V}_{0}}\gamma .......\left( 5 \right) $$

Dedifferentiate equation (4) with respect to T,

$$\dfrac{dV}{dT}={{k}^{1}}3{{T}^{2}}$$

Put values of $${{k}^{1}}$$ from equation (4)

$$ \dfrac{dV}{dT}=\dfrac{V}{{{T}^{3}}}\times 3{{T}^{2}} $$

$$\dfrac{dV}{dT}=\dfrac{3V}{T} $$

$$\therefore \dfrac{dV}{V}=3\dfrac{dT}{T} $$

$$dV=\dfrac{3}{T}VdT......(6)$$

Volume expansion is given by,

$$\Delta V=\gamma {{V}_{0}}\Delta T......\left( 7 \right)$$

Step 6: Compare the equations (6) and (7)

$$dV=\Delta V$$

$$\dfrac{3}{T}VdT = \gamma {{V}_{0}}\Delta T......\left( 7 \right)$$

$$\gamma =\dfrac{3}{T}$$

Hence the coefficient of volume expansion of the gas is $$\gamma =\dfrac{3}{T}$$.

 

$$\begin{array}{l} From\, the\, figure \\ Equation\, of\, line\, AB \\ y-{ y_{ 1 } }=\dfrac { { { y_{ 2 } }-{ y_{ 1 } } } }{ { { x_{ 2 } }-{ x_{ 1 } } } } \left( { x-{ x_{ 1 } } } \right)  \\ P-{ P_{ 0 } }=\dfrac { { 2{ P_{ 0 } }-{ p_{ 0 } } } }{ { { V_{ 0 } }2{ V_{ 0 } } } } \left( { V-2{ V_{ 0 } } } \right)  \\ =\dfrac { { -{ P_{ 0 } } } }{ { { V_{ 0 } } } } \left( { V-2{ V_{ 0 } } } \right)  \\ P=\dfrac { { -{ P_{ 0 } } } }{ { { V_{ 0 } } } } V+3{ P_{ 0 } } \\ PV=-\dfrac { { { P_{ 0 } } } }{ { { V_{ 0 } } } } V+3{ P_{ 0 } }V \\ nRT=-\dfrac { { { P_{ 0 } } } }{ { { V_{ 0 } } } } V+3{ P_{ 0 } }V \\ T=\dfrac { 1 }{ { nR } } \left( { -\dfrac { { { P_{ 0 } } } }{ { { V_{ 0 } } } } V+3{ P_{ 0 } }V } \right)  \\ \dfrac { { dT } }{ { dV } } =0\, \, \left( { for\, \max  imum\, temperature } \right)  \\ -\frac { { { P_{ 0 } } } }{ { { V_{ 0 } } } } 2V+3{ P_{ 0 } }=0 \\ -\dfrac { { { P_{ 0 } } } }{ { { V_{ 0 } } } } 2V=-3{ P_{ 0 } } \\ V=\dfrac { 3 }{ 2 } { V_{ 0 } }\, \, \left( { condition\, for\, \max  imum\, \, temperature } \right)  \\ { T_{ \max   } }=\dfrac { 1 }{ { nR } } \left( { \dfrac { { { P_{ 0 } } } }{ { { V_{ 0 } } } } \times \dfrac { 9 }{ 4 } V_{ 0 }^{ 2 }+3{ P_{ 0 } }\times \frac { 3 }{ 2 } { V_{ 0 } } } \right)  \\ =\frac { 1 }{ { nR } } \left( { -\dfrac { 9 }{ 4 } { P_{ 0 } }{ V_{ 0 } }+\dfrac { 9 }{ 2 } { P_{ 0 } }{ V_{ 0 } } } \right)  \\ =\dfrac { 9 }{ 4 } \dfrac { { { P_{ 0 } }{ V_{ 0 } } } }{ { nR } }  \end{array}$$