# When the temperature of a rigid hollow sphere containing 685 L of He gas is held at 348OC, the pressure of the gas is 1.89 x 103 kPa. How many moles of He does the sphere contain?

##### 1 Answer
Mar 4, 2018

$\text{251 mol}$.

#### Explanation:

(Assuming you meant $1.89 \times {10}^{3}$ kPa as the pressure.)

For this question, we're given:

• $P$, or pressure, which is $1.89 \times {10}^{3} k P a$. This is the same as $\frac{1.89 \times {10}^{3}}{101.325} = \text{18.65 atm}$.
• $V$, or volume, which is $\text{685 L}$.
• $T$, or temperature, which is 348 °C. This is the same as $348 + 273.15 = \text{621.15 K}$.

Assuming that helium behaves as an ideal gas, we can use the Ideal Gas Law:

$p V = n R T$

We can also rearrange this to have only $n$, or number of moles on one side:

$n = \frac{p V}{R T}$

Let's say that the value of $R$, the universal gas constant, is $\text{0.08206 L atm / K mol}$ in this case.
Plugging in all of the values, we get:

$n = \frac{p V}{R T}$
$n = \left(\text{18.65 atm" xx "685 L")/("0.08206 L atm / K mol" xx "621.15 K}\right)$
n = ("18.65" cancel("atm") xx "685" cancel("L"))/("0.08206" cancel("L atm") "/" cancel("K") mol xx 621.15 cancel("K"))
$n = \text{251 mol}$