# It is desired to increase the volume of a given weight of co2 at40degrees to three times original volume at constant pressure the temperature should be raised a) 333b) 120c) 666 degrees?

Dec 25, 2017

The answer is option $\left(c\right) = {666}^{\circ} C$

#### Explanation:

Apply Charles' Law

${V}_{1} / {T}_{1} = {V}_{2} / {T}_{2}$ at constant pressure

The initial volume is ${V}_{1} = V u$

The final volume is ${V}_{2} = 3 V u$

The initial temperature is ${T}_{1} = {40}^{\circ} C + 273 = 313 K$

The final temperature is

${T}_{2} = {V}_{2} / {V}_{1} \cdot {T}_{1} = 3 \frac{V}{V} \cdot 313 = 939 K$

Therefore,

The temperature should be raised by $= 939 - 273 = {666}^{\circ} C$

Dec 25, 2017

Using $P V = n R T$, the answer will be (c) ${666}^{\circ} \text{C}$.

#### Explanation:

According to the problem, pressure remains constants. Since the weight also is constant, the number of moles remains constant.

From the ideal gas equation, we can deduce that $P$ is directly proportional to the absolute temperature.

${V}_{2} / {V}_{1} = {T}_{2} / {T}_{1}$

$3 = {T}_{2} / \left(\left(40 + 273\right) \text{K}\right)$

So

${T}_{2} = \text{939 K}$

${T}_{2} = \text{939 K" - 273 = 666^@"C}$