A gas at a certain volume and temperature has pressure 75 cm. If the mass of the gas is doubled at the same volume and temperature, its new pressure is

To find the new pressure of the gas when its mass is doubled under constant volume and temperature, we can use the ideal gas law:

$PV=nRT$

where:
- $P$ is the pressure of the gas,
- $V$ is the volume,
- $n$ is the number of moles,
- $R$ is the universal gas constant, and
- $T$ is the absolute temperature.

The number of moles $n$ is directly proportional to the mass $m$ of the gas, given by:
$n=\frac{m}{M}$
where $M$ is the molar mass of the gas. Substituting this into the ideal gas equation yields:

$PV=\frac{mRT}{M}$

From this equation, we can express the pressure $P$ as:

$P=m\times \left(\frac{RT}{MV}\right)$

Since the volume ($V$) and temperature ($T$) are kept constant, and the molar mass ($M$) and gas constant ($R$) are inherent constants for the given gas, the term inside the parentheses remains constant. Therefore, the pressure of the gas is directly proportional to its mass:

$P\propto m$

This relationship allows us to write:
$\frac{P_2}{P_1}=\frac{m_2}{m_1}$

Given details from the question:
- Initial pressure, $P_1=75\text{cm}$
- The mass of the gas is doubled, so $m_2=2m_1$

Substitute these values into the ratio:

$\frac{P_2}{75}=\frac{2m_1}{m_1}$

Simplifying the equation gives:

$\frac{P_2}{75}=2$

Solving for $P_2$:

$P_2=75\times 2=150\text{cm}$

Therefore, the new pressure of the gas is 150 cm.