When the volume of an ideal gas is doubled while the temperature is halved, what happens to the pressure?

To understand what happens to the pressure of an ideal gas when the volume is doubled and the temperature is halved, we can apply the ideal gas law, which states:

[ PV = nRT ]

where:

• ( P ) = pressure

• ( V ) = volume

• ( n ) = number of moles of gas (constant)

• ( R ) = ideal gas constant (constant)

• ( T ) = temperature in Kelvin

If we denote the initial conditions of the gas as ( P_1 ), ( V_1 ), and ( T_1 ), and describe the new conditions as ( V_2 ) and ( T_2 ), we can establish the relationships:

1. The volume is doubled, so ( V_2 = 2V_1 ).

2. The temperature is halved, hence ( T_2 = \frac{1}{2}T_1 ).

To find the relationship of the new pressure ( P_2), we can rearrange the ideal gas law to express pressure:

[ P_1 = \frac{nRT_1}{V_1} \quad \text{and} \quad P_