Problem Analysis
We are given an ideal gas expanding under a specific condition: $P T^{3} = \text{constant}$. We need to find the gas's coefficient of volume expansion, $\beta$. The ideal gas law is $PV = nRT$. The coefficient of volume expansion is defined as $\beta = \frac{1}{V} \left( \frac{\partial V}{\partial T} \right)_P$. However, the condition $P T^{3} = \text{constant}$ implies pressure ($P$) changes with temperature ($T$). Therefore, we calculate the effective coefficient along the given process path: $\beta_{eff} = \frac{1}{V} \frac{dV}{dT}$.
Calculating the Coefficient
- Isolate Pressure (P): From the given condition $P T^{3} = K$ (where $K$ is a constant), we get: $ P = \frac{K}{T^3} $
- Substitute P into the Ideal Gas Law: Using $PV = nRT$, substitute the expression for $P$: $ \left( \frac{K}{T^3} \right) V = nRT $
- Solve for Volume (V): Rearrange the equation to express $V$ as a function of $T$: $ V = \frac{nRT^4}{K} $
- Find the Rate of Change of Volume with Temperature: Differentiate $V$ with respect to $T$ along this specific process: $ \frac{dV}{dT} = \frac{d}{dT} \left( \frac{nRT^4}{K} \right) $ $ \frac{dV}{dT} = \frac{nR}{K} \frac{d}{dT}(T^4) = \frac{nR}{K} (4T^3) = \frac{4nRT^3}{K} $
- Compute the Effective Volume Expansion Coefficient ($\beta_{eff}$): Use the formula $\beta_{eff} = \frac{1}{V} \frac{dV}{dT}$ and substitute the expressions for $V$ and $\frac{dV}{dT}$: $ \beta_{eff} = \frac{1}{\left( \frac{nRT^4}{K} \right)} \times \left( \frac{4nRT^3}{K} \right) $
- Simplify the Result: Cancel terms to find the final coefficient: $ \beta_{eff} = \frac{K}{nRT^4} \times \frac{4nRT^3}{K} = \frac{4}{T} $
Final Answer
The coefficient of volume expansion for the gas under the given condition is $\frac{4}{T}$.
Similar Questions
The strain-stress plot for materials $A, B, C$ and $D$ is shown in the figure. Which material has the largest Young's modulus ?
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($\pi = 22/7$) - The temperature of a metal strip having coefficient of linear expansion $\alpha$ is increased from $T_1$ to $T_2$ resulting in increase of its length by $\Delta L_1$. The temperature is further increased from $T_2$ to $T_3$ such that the increase in its length is $\Delta L_2$.
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(take $g = 10 \text{ m/s}^2$ and $\pi = 3.14$) - Given below are two statements:
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In the light of the above statements, choose the correct answer from the options given below A water drop of radius $1$ cm is broken into eight equal droplets. Surface tension of water is $0.075$ N $m^{-1}$. The gain in surface energy is __________$\times10^{-7}$ J.(Take $\pi = 3.14$)
A gas is kept in a container having walls which are thermally non-conducting. Initially the gas has a volume of $800 \ cm^3$ and temperature $27^{\circ}C$. The change in temperature when the gas is adiabatically compressed to $200 \ cm^3$ is:
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A wire of length 10 cm and diameter 0.5 mm is used in a bulb. The temperature of the wire is 1727°C and power radiated by the wire is 94.2 W. Its emissivity is $\frac{x}{8}$ where x =……
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Important Questions from Physical and Thermal Properties of Bulk Matter
The strain-stress plot for materials $A, B, C$ and $D$ is shown in the figure. Which material has the largest Young's modulus ?
- Eight mercury drops, each of radius $r$, coalesce to form a bigger drop. The surface energy released in this process is ___________. ($\text{S}$ is the surface tension of mercury).
- The surface tension of a soap solution is $3.5 \times 10^{-2}~\text{N/m}$. The work required to increase the radius of a soap bubble from $1~\text{cm}$ to $2~\text{cm}$ is $\alpha \times 10^{-6}~\text{J}$. The value of $\alpha$ is________.
($\pi = 22/7$) - The temperature of a metal strip having coefficient of linear expansion $\alpha$ is increased from $T_1$ to $T_2$ resulting in increase of its length by $\Delta L_1$. The temperature is further increased from $T_2$ to $T_3$ such that the increase in its length is $\Delta L_2$.
Given $T_3 + T_1 = 2T_2$ and $T_2 - T_1 = \Delta T$, the value of $\Delta L_2$ is _______. - A lift of mass 1600 kg is supported by thick iron wire. If the maximum stress which the wire can withstand is $4 \times 10^8 \text{ N/m}^2$ and its radius is 4 mm, then maximum acceleration the lift can take is _________ $\text{m/s}^2$.
(take $g = 10 \text{ m/s}^2$ and $\pi = 3.14$)
