6 public questions tagged with this topic.
Given: T1 = 27∘C = 300K, P1 = 2atm, V1 = 4L, T2 = 127∘C = 400K, P2 = 1atm. Ideal gas law: P1V1T1 = P2V2T2. V2 = V1×P1P2×T2T1 = 4×21×400300 = 4×2×43 = 10.67L.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
Given: P1 = 1.8atm, T1 = 27∘C = 300K, V1 = 3.5L, P2 = 3atm, T2 = 127∘C = 400K. P1V1T1 = P2V2T2. V2 = V1×P1P2×T2T1 = 3.5×1.83×400300 = 3.5×0.6×1.3333≈2.8L. Percentage change: V1−V2V1×100 = 3.5−2.83.5×100 = 0.73.5×100 = 20% (decrease).
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
Given: P1 = 1atm, V1 = 2L, T1 = 300K, V2 = 0.5L, T2 = 450K. P1V1T1 = P2V2T2. P2 = P1×V1V2×T2T1 = 1×20.5×450300 = 1×4×1.5 = 6atm. As per NCERT, applying relevant law/formula with correct units and sign convention leads to 6 atm. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
According to Boyle’s Law (Section 10.4), for an ideal gas at constant temperature, PV = constant. If pressure doubles (P2 = 2P1), then V2 = V1/2, so the volume halves.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
Charles’ Law states that at constant pressure, V/T = constant. If volume doubles (V2 = 2V1), the absolute temperature doubles (T2 = 2T1). As per NCERT, applying relevant law/formula with correct units and sign convention leads to It doubles. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
Boyle’s Law (Section 10.4) states PV = constant at constant temperature. If V2 = V1/2, then P2 = 2P1, so the pressure doubles. As per NCERT, applying relevant law/formula with correct units and sign convention leads to It doubles. This satisfies dimensional consistency and physical conditions given, so option C is scientifically correct.
Ref: NCBI Bookshelf, Ideal Gas Law: P1V1/T1 = P2V2/T2.
