Question

A sample of argon gas is sealed in a container. The volume of the container is doubled. If the
pressure remains constant, what happens to the absolute temperature?

Answer 1

A sample of argon gas is sealed in a container. The volume of the container is doubled. If the pressure remains constant, what happens to the absolute temperature?

Solution.

$$V _ {2} = 2 V _ {1}, P _ {2} = P _ {1};$$ $$T _ {1} - ?$$

Clapeyron Equation:

$$P _ {1} V _ {1} = \frac {m _ {1}}{M} R T _ {1}.$$

$P_{1}$ - the pressure of the argon gas;

$V_{1}$ - the volume the container;

$m_{1}$ - the mass of the argon gas;

$M$ - the molar mass of the argon gas;

$R$ - the gas constant;

$T_{1}$ - the absolute temperature.

Clapeyron Equation after doubling of the volume:

$$P _ {2} V _ {2} = \frac {m _ {2}}{M} R T _ {2};$$

A sample of argon gas is sealed then the mass is the same:

$$m _ {2} = m _ {1}.$$

The pressure remains constant:

$$P _ {2} = P _ {1}.$$

The volume of the container is doubled:

$$V _ {2} = 2 V _ {1}.$$ $$P _ {1} 2 V _ {1} = \frac {m _ {1}}{M} R T _ {2}.$$

First equation: $P_{1}V_{1} = \frac{m_{1}}{M} RT_{1}$ .

Second equation: $P_{1}2V_{1} = \frac{m_{1}}{M} RT_{2}$ .

Divide second equation by first equation:

$$\frac {T _ {2}}{T _ {1}} = 2;$$ $$T _ {2} = 2 T _ {1}.$$

Answer: The absolute temperature is doubled: $T_{2} = 2T_{1}$.