An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and a is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by

Question

An external pressure P is applied on a cube at 0°C so that it is equally compressed from all sides. K is the bulk modulus of the material of the cube and a is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by (A) (3α/PK) (B) 3PKα (C) (P/3α K) (D) (P/α K)

Tardigrade Admin · Accepted Answer

Due to thermal expansion (Δ V/V) = 3α Δ T ....(1) Volumetric strain due to external pressure (Δ V/V) = (P/K) .....(2) Equating both equations we get 3α Δ T = (P/K) ⇒ Δ T = (P/3α K)

$\frac{3 α}{P K}$

$3 P K α$

$\frac{P}{3 α K}$

$\frac{P}{α K}$

Due to thermal expansion
$\frac{Δ V}{V} = 3 α Δ T .... (1)$
Volumetric strain due to external pressure
$\frac{Δ V}{V} = \frac{P}{K} ..... (2)$
Equating both equations we get
$3 α Δ T = \frac{P}{K}$
$\Rightarrow Δ T = \frac{P}{3 α K}$

PK3α​

3PKα

3αKP​

αKP​

Due to thermal expansion VΔV​=3αΔT....(1) Volumetric strain due to external pressure VΔV​=KP​.....(2) Equating both equations we get 3αΔT=KP​ ⇒ΔT=3αKP​

$\frac{3 α}{P K}$

$3 P K α$

$\frac{P}{3 α K}$

$\frac{P}{α K}$

Due to thermal expansion
$\frac{Δ V}{V} = 3 α Δ T .... (1)$
Volumetric strain due to external pressure
$\frac{Δ V}{V} = \frac{P}{K} ..... (2)$
Equating both equations we get
$3 α Δ T = \frac{P}{K}$
$\Rightarrow Δ T = \frac{P}{3 α K}$