Question

If 3.0L of helium at 20.0 is allowed to expand to 4.4L with the pressure remaining the same what is the new temp:?
1.702C
2.430C
3.157C
4.30.0C​

Answer 1

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Answer 2

The answer is option 3. 157 °C.

Explanation:

Given:

• Initial Volume $V_1 = 3.0 L$
• Final Volume $V_2 = 4.4 L$
• Initial Temperature $T_1 = 20 \°C = 20+273 = 293 K$

To find: Final Temperature $T_2$

Formula to be used: Charle's Law $\frac{V_1}{T_1} =\frac{V_2}{T_2}$

Solution:

Substituting the values in the formula

$\frac{3}{293} =\frac{4.4}{T_2}$

$T_2 =\frac{4.4\times 293}{3}$

$T_2 = 429.7 K$

But, the answer is in Kelvin. So, we would change it to degree Celsius by subtracting 273 from the answer.

$T_2 = 429.7-273 = 156.7 = 157\°C$

Hence, the final temperature would be 157 degrees celsius