1. Let the hydrocarbon be $C_xH_y$.

2. Given volumes at same $T$ and $P$, so volume ratios = mole ratios.

   • Hydrocarbon volume $= 15\,\text{mL}$
   • Air used $= 375\,\text{mL}$
   • Since air contains $20\%$ $O_2$, oxygen volume supplied is $375 \times \frac{20}{100} = 75\,\text{mL}$

3. Combustion reaction: $C_xH_y + \left(x + \frac{y}{4}\right)O_2 \rightarrow xCO_2 + \frac{y}{2}H_2O$

   Since water is liquid, it does not contribute to final gas volume.

4. Use oxygen consumption data

   For $15\,\text{mL}$ hydrocarbon, oxygen required is $75\,\text{mL}$.

   So for $1$ volume hydrocarbon, oxygen required is $\frac{75}{15} = 5 \text{ volumes}$

   Hence, $x + \frac{y}{4} = 5$ $4x + y = 20 \quad ...(1)$

5. Final gaseous volume after combustion

   Initial gases present before combustion:

   • Hydrocarbon = $15$ mL
   • $O_2 = 75$ mL
   • $N_2$ from air = $375 - 75 = 300$ mL

   After combustion:

   • $CO_2 = 15x$ mL
   • Unreacted $O_2 = 0$ mL (since required exactly for complete combustion)
   • $N_2 = 300$ mL
   • Water is liquid, so ignored in gas volume

   Total final gas volume is given as $345$ mL: $15x + 300 = 345$ $15x = 45$ $x = 3$

6. Find $y$ using equation (1): $4x + y = 20$ $4(3) + y = 20$ $12 + y = 20$ $y = 8$

7. Therefore, the hydrocarbon is $C_3H_8$

8. Check with options

   • A: $C_3H_8$ ✅
   • B: $C_4H_8$ ❌
   • C: $C_4H_{10}$ ❌
   • D: $C_3H_6$ ❌

So the correct option is A.