If a gas follows Boyle's law then what will be the form of the graph between PV and V?
- A
- B
- C
- ✓
Answer: D.
View full solution →140 questions across 9 question groups — pick any mix to generate a Physics paper with step-by-step answer keys.
M.C.Q (1 Marks)
23 Q→02Fill In The Blanks[1 Marks ]
9 Q→03True False[1 Marks ]
10 Q→041 Marks Question
44 Q→052 Marks Questions
29 Q→063 Marks Question
4 Q→074 Marks Question
2 Q→08Match the following.
2 Q→095 Marks Questions
17 Q→One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
Answer: D.
View full solution →Answer: C.
View full solution →Answer: C.
View full solution →Answer: C.
View full solution →Answer: C.
View full solution →| A | B |
| 1. Any gas behave at high temperature and low pressure | (A) $\frac{89}{27 b^2}$ |
| 2. The value of $N_A$ will be | (B) $6.023 \times 10^{26}$ per kg mol. |
| 3. $P _{ C }=$ | (C) Like ideal gas |
| 4. $T _{ C }=$ | (D)$\frac{a}{27 b^2}$ |
| 5. In a container the mean free path of molecule is inversely proportional to | (E) Density |
| A | B |
| 1. Dalton's Partial Pressure law | (A) $\frac{2}{3}$ |
| 2. PV = | (B) Excessive |
| 3. The part of mean K.E of unit volume gas isequal to | (C) $\begin{array}{l} P = P _1+ P _2+ P _3+\ldots\end{array}$ |
| 4. The value of KE. due to r.m.s. velocity is zero | (D) $\frac{1}{3} MC _{r m s}^2$ |
| 5. The gas that's density ( $\rho$ ) become less than value of $C _{ rms }$ | (E) zero |

Pick question groups from the list above, set marks and difficulty, and export a branded PDF with step-by-step answer keys. First 3 chapters free — no signup.