Question

A balloon is filled with hydrogen at room temperature. It will burst if the pressure exceeds $5\text{bar}$. If at $1\text{bar}$ pressure the gas occupies $2.5\text{L}$, upto what volume can the balloon be compressed?

• A. $0.5\text{L}$
• B. $1.2\text{L}$
• C. $1.5\text{L}$
• D. $0.8\text{L}$

Answer 1

The correct option is A $0.5\text{L}$

According to Boyle’s Law, $P_1{V}_1=P_2{V}_2$
If $P_1$ is 1 bar, $V_1$ will be $2.5\text{L}$
If $P_2=5$ bar, then $V_2=\frac{P_1{V}_1}{P_2}=\frac{1\times 2.5}{5}\text{L}=0.5\text{L}$
Since the balloon bursts above $5\text{bar}$ pressure, the volume of the balloon should be more than $0.5\text{L}$ to avoid bursting.
Theory:

Gas Laws: Gas laws give the relationship between the different measurable quantities (pressure, volume, temperature, and amount) of gases.
If these four quantities form any equation, that equation is called an equation of state.
Units of Pressure $\left(P\right)$:
$1\text{atm}=1.01325\times {10}^5\text{Pa}=101.325\text{kPa}$
$1\text{bar}={10}^5\text{Pa}=750\text{torr}=750\text{mm of Hg}$
$1\text{atm}=760\text{torr}=760\text{mm of Hg}=76\text{cm of Hg}$
$1\text{atm}=1.01325\text{bar}$
$1{\text{N/m}}^2=1\text{Pa}=10{\text{dyne/cm}}^2$
Units of Volume $\left(V\right)$:
$1{\text{m}}^3={10}^3{\text{dm}}^3\text{or L}$
$1\text{L}={10}^3{\text{cm}}^3\text{or c.c. or mL}$
$1{\text{m}}^3={10}^6{\text{cm}}^3$
Unit of Temperature $\left(T\right)$:
$T\left(\text{K}\right)=t{(}^{\circ }\text{C})+273.15\approx t{(}^{\circ }\text{C})+273$
Note: Always use K as the temperature unit.
Standard Temperature and Pressure(STP) = Normal Temperature and Pressure(NTP):
STP (old or atm)
$T=273.15\text{K}$
$P=1\text{atm}$
$\text{Molar volume of an ideal gas}=22.4{\text{L~mol}}^{-1}$
STP (new or bar)
$T=273.15\text{K}$
$P=1\text{bar}$
$\text{Molar volume of an ideal gas}=22.7{\text{L~mol}}^{-1}$