A gas is initially at P1 = 2 atm and V1 = 6 L. If P2 = 3 atm, what is V2 for an isothermal process?

A gas is initially at P1 = 2 atm and V1 = 6 L. If P2 = 3 atm, what is V2 for an isothermal process?

• A. 2 L
• B. 4 L
• C. 6 L
• D. 8 L

Answer: (B)

Isothermal means the temperature stays the same, so for an ideal gas the product PV remains constant (PV = nRT with T fixed). This gives P1V1 = P2V2. Here P1V1 = 2 atm × 6 L = 12 atm·L. With P2 = 3 atm, V2 = (P1V1)/P2 = 12 atm·L / 3 atm = 4 L. So the volume must drop from 6 L to 4 L when the pressure rises from 2 atm to 3 atm in an isothermal process. The other volumes don’t keep PV constant (3 atm × 2 L = 6, 3 atm × 6 L = 18, 3 atm × 8 L = 24), whereas 3 atm × 4 L = 12 matches P1V1.

Isothermal means the temperature stays the same, so for an ideal gas the product PV remains constant (PV = nRT with T fixed). This gives P1V1 = P2V2. Here P1V1 = 2 atm × 6 L = 12 atm·L. With P2 = 3 atm, V2 = (P1V1)/P2 = 12 atm·L / 3 atm = 4 L. So the volume must drop from 6 L to 4 L when the pressure rises from 2 atm to 3 atm in an isothermal process. The other volumes don’t keep PV constant (3 atm × 2 L = 6, 3 atm × 6 L = 18, 3 atm × 8 L = 24), whereas 3 atm × 4 L = 12 matches P1V1.