A hot air balloon is filled with $1.89 x 10^2$ liters of air at 21 C. If atmospheric pressure does not change, how hot must the air become in order to increase the volume to $4.5 x 10^2$ liters?

As the initial volume of 20 L is less than the final volume of 24 L, the temperature must increase. This is based on Charles' Law, which states that there...

$V_1/T_1=V_2/T_2$; this assumes a given quantity of gas, and absolute temperatures...... And so....$T_2=V_2/V_1xxT_1=(5.70*L)/(1.90*L)xx(21.0+273.13*K)$ $=??*K$

So, we need first to calculate the mass of the helium in the balloon. We are told that there are $1.2*L$ of $He$, and also that the ($rho$) of $He$...

Given $n="Mass"/"Molar mass"=(PV)/(RT)$. $"Mass"=(PV)/(RT)xx"molar mass"$, .........Agreed........? And, as is typical, we use the value of.......... $R=0.0821*L*atm*K^-1*mol^-1$ The which has units that chemists typically use. And so we plugs in the...

$V=(10^3*molxx0.0821*L*atm*K^-1*mol^-1xx298*K)/(1*atm)$ Note that $101*kPa~=1*atm$ We get $V=24.4xx10^3L=24.4*m^3$. What is the mass of the gas inside the balloon? Will it float in air? This problem reminds me of an old...

This is a Charles Law problem $ V_1/T_1 = V_2/T_2$ $ V_1 = "3.5 L"$, $ T_1 = 25 + 273 = "298 K"$ , $V_2 =...

Since, for a given quantity of gas, $VpropT$ if $P$ is constant, $V=kT$. Solving for $k$, $V_1/T_1=V_2/T_2$, and $V_2=(V_1xxT_2)/T_1$; $T$ is reported in $"degrees Kelvin"$, $V$ may be in whatever...

This is an example of , which states that the volume of a gas held at constant pressure varies directly with the Kelvin temperature. The equation for this law is:...

At first glance, this appears to be an problem. However, water is a liquid at 24 °C and 2.27 atm. For argument's sake, let's say that the temperature on...

We use old $"Boyle's Law"$, $P_1V_1=P_2V_2$ at constant temperature. And we solve for $P_2=(P_1V_1)/V_2=(195*Lxx0.993*atm)/(7.5xx10^3L)$ $=2.58xx10^-2*atm.$