Question

10.The volume V of a given mass of gas varies inversely as the pressure P. When V =2m3 ,P=500 N/m2 .Find the volume when the pressure is 400N/m2 Find the pressure when the volume is 5 m3

Answer 1

Answer:

$2.5\; \rm m^3$.

Step-by-step explanation:

Let $V_1$ and $P_1$ denote the volume and pressure of this gas before the change. Let $V_2$ and $P_2$denote the pressure of this gas after the change.

The ratio between the volume of this gas before and after the change would thus be $\displaystyle \frac{V_1}{V_2}$.

The ratio between the pressure of this gas before and after this change would be $\displaystyle \frac{P_1}{P_2}$.

Because the volume of a given mass of (ideal) gas varies inversely with pressure, these two ratios should be reciprocal of each other. That is:

$\displaystyle \frac{V_1}{V_2} = \frac{1}{\displaystyle \displaystyle \left(\frac{P_1}{P_2}\right)}$.

In other words:

$\displaystyle \frac{V_1}{V_2} = \frac{P_2}{P_1}$.

Rearrange and solve for $V_2$, the volume of this gas after the change:

$\displaystyle V_1 = V_2\cdot \left(\frac{P_2}{P_1}\right)$.

$V_2 = \displaystyle V_1 \cdot \left(\frac{P_1}{P_2}\right)$.

From the question:

• Volume of this gas before the change: $V_1 = 2\; \rm m^3$.
• Pressure of this gas before the change: $P_1 = 500\; \rm N \cdot m^{-2}$.
• Pressure of this gas after the change: $P_2 = 400\; \rm N \cdot m^{-2}$.

Solve for $V_2$:

$\begin{aligned} V_2 &= V_1 \cdot \left(\frac{P_1}{P_2}\right) \\ &= 2\; \rm m^3 \times \frac{500\; \rm N \cdot m^{-2}}{400\; \rm N \cdot m^{-2}} = 2.5\; \rm m^3\end{aligned}$.