The volume of 1 gram of methane,ethane,propane, and butane was measured at 350 K and 1 atmosphere .What is the volume of butane?

Open in App

Solution

$Use V = \frac{n R T}{P} t o s o l v e n = \frac{1}{58} = 0.017241 moles V = \frac{0.017241 \times 0.0821 \times 350}{1} = 0.5 L$

Suggest Corrections

Similar questions

Q. The volume of

1 g

each of methane

(C H_{4})

, ethane

(C_{2} H_{6})

, propane

(C_{3} H_{8})

and butane

(C_{4} H_{10})

was measured at

350 K

and

1 a t m

. What is the volume of butane ?

Q. The volume of 1 g each of methane (

C H_{4}

), ethane(

C_{2} H_{2}

) , propane (

C_{3} H_{8}

) and butane (

C_{4} H_{10}

) was measured at 350 K and 1atm. What is the volume of butane

The volume of 1 g each of methane ( $C H_{4}$ ), ethane ( $C_{2} H_{6}$ ) , propane ( $C_{4} H_{1} 0$ ) was measured at 350 K and 1 atm. What is the volume of butane

The main constituent of petroleum gas is :
(a) methane (b) ethane (c) butane (d) propane

Which is the major component of CNG ?

(a) ethane

(b) propane

(d) butane

Join BYJU'S Learning Program

The volume of 1 gram of methane,ethane,propane, and butane was measured at 350 K and 1 atmosphere .What is the volume of butane?

Use V = nRTP to solven=158=0.017241molesV = 0.017241 × 0.0821 × 3501=0.5L

$Use V = \frac{n R T}{P} t o s o l v e n = \frac{1}{58} = 0.017241 moles V = \frac{0.017241 \times 0.0821 \times 350}{1} = 0.5 L$