Question

The temperature of a certain mass of gas was increased from ${40}^{0}C$ to ${41}^{0}C$ at constant pressure. The volume of the gas?

• A. will remain constant
• B. will increase by $\frac{1}{273}$ of its volume at 273 K
• C. will increase by $\frac{1}{273}$ of its volume at ${40}^{0}C$
• D. will increase, but the increase in volume cannot be predicted.

Answer 1

Answer: (C)

will increase by $\frac{1}{273}$ of its volume at ${40}^{0}C$

Answer : B . will increase by $\frac{1}{273}$ of its volume at $273K$

From Charle's law, at constant pressure, $\text{V}\propto \text{T}$

Let us take,

$V_1$ - volume at ${40}^{o}C$

$V_2$ - volume at ${41}^{o}C$

$T_1$ = ${40}^{o}C=40+273K=313K$

$T_2$ = ${41}^{o}C=41+273K=314K$

Then,

$\frac{V_2}{V_1}=\frac{T_2}{T_1}=\frac{314}{313}$

$=>V_2=\frac{314}{313}{V}_1$

So, increase in volume is.

$V_2-V_1=\frac{314}{313}{V}_1-V_1=(\frac{314}{313}-1)V_1=\frac{1}{313}{V}_1$ ---------> ( 1 )

That is, volume will increase by, $\frac{1}{313}$ of $V_1$ volume at ${40}^{o}C$. Therefore option C in INCORRECT.

To check whether option B is correct.

Let us take,

$V_3$ - volume at $273K$

$T_3$ - $273K$

Therefore, $\frac{V_1}{V_3}=\frac{T_1}{T_3}=\frac{313}{273}$

$=>V_1=\frac{313}{273}{V}_3$ -----------------> ( 2 )

Substitute equation ( 2 ) in ( 1 ) we get,

Increase in volume, $V_2-V_1=\frac{1}{313}\times \frac{313}{273}{V}_3=\frac{1}{273}{V}_3$

That is, volume will increase by, $\frac{1}{273}$ of $V_3$ - volume at $273K$. Therefore option B in CORRECT.