Question

A 0.500-mol sample of an ideal monatomic gas at 400 kPa and 300 K, expands quasi-statically until the pressure decreases to 160 kPa. Find the final temperature and volume of the gas, the work done by the gas, and the heat absorbed by the gas if the expansion is (a) isothermal, and (b) adiabatic

Answer 1

Answer:

(a). The final volume is 2.5 times the initial volume.

The work done is 1143 J.

(b). The final temperature is 207.9 K

The work done is 574 J.

Explanation:

Given that,

Sample of an ideal gas = 0.500 mol

Initial pressure = 400 kPa

Final pressure = 160 kPa

Temperature = 300 K

(a) for isothermal,

Temperature will be same.

We need to calculate the volume of gas

$P_{f}V_{f}=P_{i}V_{i}$

$V_{f}=(\dfrac{P_{i}}{P_{f}})V_{i}$

Put the value into the fomrula

$V_{f}=(\dfrac{400}{160})V_{i}$

$V_{f}=2.5 V_{i}$

We need to calculate the work done

Using equation of energy

$dQ=dW$

$dQ=nRTln(\dfrac{P_{in}}{P_{f}})$

$dQ=0.500\times8.314\times300\times ln(\dfrac{400}{160})$

$dQ=1143\ J$

(b). For adiabatic,

No transfer of heat between system and surroundings

We need to calculate the final temperature

Using formula of gas

$P_{f}^{1-\gamma}T_{f}^{\gamma}=P_{i}^{1-\gamma}T^{\gamma}$

$T_{f}=(\dfrac{P_{i}}{P_{f}})^{\frac{1-\gamma}{\gamma}}T_{i}$

Put the value into the formula

$T_{f}=(\dfrac{400}{160})^{\frac{1-\frac{5}{3}}{\frac{5}{3}}}\times300$

$T_{f}=207.9\ K$

We need to calculate the wok done in adiabatic

Using formula of work done

$W=\dfrac{nR(T_{i}-T_{f})}{\gamma-1}$

$W=\dfrac{0.500\times8.314\times(300-207.9)}{\dfrac{5}{3}-1}$

$W=574\ J$

Hence, (a). The final volume is 2.5 times the initial volume.

The work done is 1143 J.

(b). The final temperature is 207.9 K

The work done is 574 J.