Question

a sample of ideal gas is originally at 40 degrees celsius. what is the final temperature of the gas if the volume is doubled and pressure is reduced to half of the original values?

Answer 1

Answer:

313 K or 40 degree celsius.

Explanation:

Initial , $V_{i}=V_{0}$

$P_{i}=P_{0}$

$T_{i} =40+273=313K$

Now, final

$P_{f} =\frac{1}{2}P_{0}$

$V_{f} =2V_{0}$

Combined  the ideal gas law

$\frac{P_{i}V_{i}}{T_{i}}=\frac{P_{f}V_{f}}{T_{f}} \\\frac{P_{0}V_{0}}{T_{0}}=\frac{1}{2} \frac{P_{0}2V_{0}}{T_{f}} \\T_{f}=T_{0}\\T_{f}=313K$

Therefore, the final temperature of ideal gas is 313 K.