#gas pressure

Latent Heat and Phase Transitions

A gas at 37∘C and 3atm occupies 6L. If it is cooled to −13∘C while the volume is reduced to 4L, what is the final pressu

Given: T1 = 37∘C = 310K, P1 = 3atm, V1 = 6L, T2 = −13∘C = 260K, V2 = 4L. Ideal gas law: P1V1T1 = P2V2T2. P2 = P1×V1V2×T2T1 = 3×64×260310. P2 = 3×1.5×0.8387≈3.77atm.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Specific Heat and Calorimetry

A gas at 2.8atm and 97∘C occupies 3.8L. If the temperature is raised to 147∘C and volume reduced to 2.5L, what is the fi

Given: P1 = 2.8atm, T1 = 97∘C = 370K, V1 = 3.8L, T2 = 147∘C = 420K, V2 = 2.5L. P1V1T1 = P2V2T2. P2 = P1×V1V2×T2T1 = 2.8×3.82.5×420370. P2 = 2.8×1.52×1.1351≈4.83atm.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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Specific Heat and Calorimetry

A gas at 300K and 1atm occupies 3L. If it is heated to 450K while the volume is adjusted to 4.5L, what is the final pres

Given: T1 = 300K, P1 = 1atm, V1 = 3L, T2 = 450K, V2 = 4.5L. P1V1T1 = P2V2T2. P2 = P1×V1V2×T2T1 = 1×34.5×450300 = 1×23×1.5 = 1atm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 1 atm. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.

Ref: NCERT Class 11 Physics, Chapter 11: Thermal Properties – Calorimetry.

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Temperature Scales and Triple Point

A gas at 3.5atm and 77∘C occupies 4.5L. If the temperature drops to 27∘C and volume increases to 6L, what is the final p

Given: P1 = 3.5atm, T1 = 77∘C = 350K, V1 = 4.5L, T2 = 27∘C = 300K, V2 = 6L. P1V1T1 = P2V2T2. P2 = P1×V1V2×T2T1 = 3.5×4.56×300350 = 3.5×0.75×0.8571≈2.25atm.

Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.

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