Question

A sample of neon gas in a closed vessel occupies 85.0 mL at 25.0 °C. What is its new volume, in mL, if the temperature decreases to –16.0 °C, with P and n constant?

Answer 1

Answer:

The new volume of the gas is 73.31 mL at -16°C.

Explanation:

Boyle's Law:

The pressure of a given mass at a constant temperature of an ideal gas is inversely proportion to its volume.

$P\propto \frac 1V$

Charles' Law:

The volume is directly proportional to the temperature of an ideal gas of a given mass at a constant pressure.

$V\propto T$

Combined two gas laws

$PV\propto T$

$\therefore \frac{P_1V_1}{T_1}= \frac{P_2V_2}{T_2}$

Given that,

A sample of neon gas closed vessel occupied 85.0 mL at 25.0°C with constant P and n.

Here $P_1=P_2$, $V_1=85.0$ mL, $T_1$ =( 25+273)K=298 k

$V_2=$? , $T_2$=(-16+273)K=257 k

Since the pressure is constant.

So, the gas equation becomes

$\therefore \frac{V_1}{T_1}=\frac{V_2}{T_2}$

Putting the value of $V_1$,$T_1$ and $T_2$

$\Rightarrow\frac{85.0}{298}=\frac{V_2}{257}$

$\Rightarrow V_2=\frac{85.0\times 257}{298}$

$\Rightarrow V_2= 73.31$ mL

The new volume of the gas is 73.31 mL at -16°C.