Assertion :In a gas any rapid change must be adiabatic whereas a slow change may be adiabatic. Reason: In a P-V diagram, the magnitude of the slope is greater for an adiabatic process than for an isothermal process.

Both statements 1 and 2 are true and statement 2 is the correct explanation of statement 1.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Both statements 1 and 2 are true but statement 2 is not the correct explanation of statement 1.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Statement 1 is true but statement 2 is false.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Statement 1 is false but statement 2 is true.

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D Statement 1 is false but statement 2 is true.
For adiabatic process, change in heat energy is zero so, process occurs rapidly or suddenly like burst of a tyre etc. but when process occurs slowly there is heat exchange and process become isothermal not adiabatic and

$\Rightarrow \frac{S l o p e o f a d i a b a t i c}{S l o p e o f i s o t h e r m a l} = γ .$
Hence, the answer is statement 1 is false and statement 2 is true.

Suggest Corrections

Similar questions

C - O

C - O

Choose the correct option.

Statement $1 : (p \land \sim q) \land (\sim p \land q)$ is a fallacy.

Statement $2 : (p \Rightarrow q) \Leftrightarrow (\sim q \Rightarrow\sim p)$ is a tautology.

Statement 1: A polynomial P(x) of degree greater than or equal to one, when divided by x – a, leaves a remainder zero.

Statement 2: x – a is a factor of the polynomial P(x).

1

\to A = 2^i + 3^j + 6^k, \to B =^i +^j - 2^k

\to C =^i + 2^j +^k

| \to A \times (\to A \times (\to A \times \to B)) \cdot \to C | = 243

2

| \to A \times (\to A \times (\to A \times \to B)) \cdot \to C | = | \to A |^{2} | [\to A \to B \to C] |

2 m g