The pressure and temperature of 4 $d m^{3}$ of $C O_{2}$ gas are doubled. Then the volume of $C O_{2}$ gas would be

2 d m^{3}

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

3 d m^{3}

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

4 d m^{3}

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

8 d m^{3}

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

Open in App

Solution

The correct option is C
$4 d m^{3}$
$C O_{2} g a s V_{1} = 4 d m^{3}; V_{2} = ? P_{1} = P; P_{2} = 2 P T_{1} = T; T_{2} = 2 T Ideal gas equation \frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}} \frac{P \times 4}{T} = \frac{2 P \times v_{2}}{2 T} 4 d m^{3} = V_{2}$

Suggest Corrections

Similar questions

4 d m^{3}

2 l i t e r s

2 L

{d m}^{3}

If pressure becomes double at the same absolute temperature on 2 L of $C O_{2}$ , then the final volume of $C O_{2}$ will be?

Join BYJU'S Learning Program