What happens to the pressure 'P' of an ideal gas, if the temperature is increased by a factor of 2 and the volume is increased by a factor of 8?
Correct answer: B. P decreases by a factor of 4
- A. P decreases by a factor of 16
- B. P decreases by a factor of 4
- C. P decreases by a factor of 2
- D. P increases by a factor of 4
- E. P increases by a factor of 16
Explanation
From PV = nRT, the pressure is P = nRT/V. Multiplying the temperature by 2 doubles the pressure while multiplying the volume by 8 divides it by 8, a net factor of 2/8 = 1/4. The pressure decreases by a factor of 4.
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Thermal equilibrium and heat explain how temperature and energy transfer are related, while molar specific heats describe the heat required by a gas at constant volume or constant pressure. The first law connects heat supplied, work done and change in internal energy, with sign conventions kept distinct from temperature changes.
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