Question

If both the pressure and volume of a given sample of an ideal gas are doubled, what happens to the temperature of the gas in Kelvins? a)The temperature of the gas in increased by four times its original value. b)The temperature of the gas is reduced to one-half its original value. c)The temperature of the gas is reduced to one-fourth its original value. d) The temperature remains constant. e)The temperature of the gas is increased by two times its original value.

Answer 1

Answer:

a)The temperature of the gas in increased by four times its original value.

Explanation:

For a fixed amount of a gas, the ideal gas equation can be written as follows:

$\frac{pV}{T}=const.$

where

p is the gas pressure

V is the gas volume

T is the gas temperature (in Kelvins)

For a gas under transformation, the equation can also be rewritten as

$\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}$

where the labels 1,2 refer to the initial and final conditions of the gas.

In this problem, we have:

- The pressure of the gas is doubled: $p_2 = 2p_1$

- The volume of the gas is doubled: $V_2=2V_1$

Substituting into the equation, we find what happens to the temperature:

$\frac{p_1V_1}{T_1}=\frac{(2 p_1)(2V_1)}{T_2}$

$\frac{1}{T_1}=\frac{4}{T_2}$

$T_2 = 4T_1$

So, the correct choice is

a)The temperature of the gas in increased by four times its original value.