A fixed mass of gas at constant pressure occupies a volume V. The gas undergoes a rise in temperature so that the root mean square velocity of molecules is doubled. The new volume will be

Question

A fixed mass of gas at constant pressure occupies a volume V. The gas undergoes a rise in temperature so that the root mean square velocity of molecules is doubled. The new volume will be (A)  (V /2) (B)  (V /√ 2) (C) 2 V (D) 4 V.

Tardigrade Admin · Accepted Answer

Using c = √T , we get, (c2/c1) = √(T2/T1) Then (2c/c) = √(T2/T) i.e. 4 = (T2/T) i.e. T2 = 4T Since V propto T, so new volume is four times.

Q. A fixed mass of gas at constant pressure occupies a volume $V$ . The gas undergoes a rise in temperature so that the root mean square velocity of molecules is doubled. The new volume will be

$\frac{V}{2}$

$\frac{V}{2}$

2 V

4 V.

Using $c = T$ , we get, $\frac{c _{2}}{c _{1}} = \frac{T _{2}}{T _{1}}$
Then $\frac{2 c}{c} = \frac{T _{2}}{T}$ $i . e .$ 4 = $\frac{T _{2}}{T}$
$i . e .$ $T_{2}$ = 4T
Since V $\propto$ T, so new volume is four times.

Q. A fixed mass of gas at constant pressure occupies a volume V. The gas undergoes a rise in temperature so that the root mean square velocity of molecules is doubled. The new volume will be

2V​

2​V​

2 V

4 V.

Using c=T​ , we get, c1​c2​​=T1​T2​​​ Then c2c​=TT2​​​ i.e. 4 = TT2​​ i.e. T2​ = 4T Since V ∝ T, so new volume is four times.

Q. A fixed mass of gas at constant pressure occupies a volume $V$ . The gas undergoes a rise in temperature so that the root mean square velocity of molecules is doubled. The new volume will be

$\frac{V}{2}$

$\frac{V}{2}$

Using $c = T$ , we get, $\frac{c _{2}}{c _{1}} = \frac{T _{2}}{T _{1}}$
Then $\frac{2 c}{c} = \frac{T _{2}}{T}$ $i . e .$ 4 = $\frac{T _{2}}{T}$
$i . e .$ $T_{2}$ = 4T
Since V $\propto$ T, so new volume is four times.