Question

One mole of oxygen at NTP is compressed
adiabatically to 5 atmospheres. What is the new
temperature and the work done ? Given : y = 1.4 and
R=831 J moleK.
Ans. 159°C, - 3299 J.​

Answer 1

Answer:

The new  temperature of gas is 159 degree C and the work done by the gas is -3300 J

Explanation:

As we know by process equation of adiabatic process is given as

$\frac{T_1^{\gamma}}{P_1^{\gamma - 1}} = \frac{T_2^{\gamma}}{P_2^{\gamma - 1}}$

now we have

$P_1 = 1 atm$

$P_2 = 5 atm$

$\gamma = 1.4$

$T_1 = 273 K$

so we have

$\frac{273^{1.4}}{1^{0.4}} = \frac{T_2^{1.4}}{5^{0.4}}$

so we have

$T_2 = (\frac{5}{1})^{0.4/1.4}(273)$

$T_2 = 432 K$

$T_2 = 159 ^oC$

now work done in this process is given as

$W = \frac{nR(T_1 - T_2)}{\gamma - 1}$

$W = \frac{1 R(0 - 159)}{1.4 - 1}$

$W = -3300 J$

#Learn

Topic : Adiabatic Process

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