Question

When the absolute temperature of a gas is doubled then the correct statements are
a) The V of a gas increases by 4 times at constant P
b) The P of a gas increases by 2 times at constant V
c) The V of a gas increases by 2 times at constant P
d) The P of a gas increases by 4 times at constant V
A. b, d
B. a, c
C. b, c
D. a, d

Answer 1

The relation between the temperature, pressure and volume can be easily expressed by Boyle's law and it is as follows.
PV = nRT
Here, P = Pressure of the gas
V = volume of the gas
n = number of moles
R = gas constant
T = Temperature of the gas

Complete answer:
- In the question it is asked to find the correct statements in the question when the temperature of the gas is doubled.
- First, we have to calculate the change in volume by keeping the pressure constant when the temperature is doubled and it is as follows.
PV = nRT
- At constant pressure, $\dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{{{T}_{1}}}{{{T}_{2}}}$
- Therefore,

& \dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{{{T}_{1}}}{{{T}_{2}}} \\\ & \dfrac{{{V}_{1}}}{{{V}_{2}}}=\dfrac{T}{2T} \\\ \end{aligned}$$ \- Means the volume of the gas is going to be doubled. \- We have to see another possibility by keeping the volume constant. \- At constant volume, $\dfrac{{{P}_{1}}}{{{P}_{2}}}=\dfrac{{{T}_{1}}}{{{T}_{2}}}$ \- Therefore, $$\begin{aligned} & \dfrac{{{P}_{1}}}{{{P}_{2}}}=\dfrac{{{T}_{1}}}{{{T}_{2}}} \\\ & \dfrac{{{P}_{1}}}{{{P}_{2}}}=\dfrac{T}{2T} \\\ \end{aligned}$$ \- Means the pressure of the gas is going to be doubled. **Therefore, the correct option is C.** **Note:** The temperature is directly proportional to the volume and the pressure. Therefore, the increase in the temperature of the gas causes the increases in the volume of the gas and pressure of the given gas respectively.