Question

A sealed tank contains 29 moles of an ideal gas at an initial temperature of The pressure of the gas is increased until the final pressure equals 1.9 times the initial pressure. The heat capacity at constant pressure of the gas is What is the heat absorbed by the gas? Let the ideal-gas constant R = 8.314 J/(mol • K).

7.0 kJ

170 kJ

230 kJ

110 kJ

-52 kJ

Answer 1

During the process approximately 110 KJ heat will be absorbed by the gas.

Answer: Option D (110KJ)

Explanation:

According to the question, the given values are,

The initial temperature,

$T_{i}=270 K$

Let us assume that the final temperature will be $T_{f}$, the initial pressure will be $P_{i}$ and the final pressure will be,

$P_{f}=1.9 P_{i}$

As we know that the Ideal Gas Equation at the constant volume is,

$P V=n R T$

Then, for initial conditions,

$P_{i} V=n R T_{i}$

And, for the final conditions,

$P_{f} V=n R T_{f}$

Now, on dividing both the equations, we get;

$\frac{P_{i}}{P_{f}}=\frac{T_{i}}{T_{f}}$

By putting all the values, we get,

$\frac{P_{i}}{1.9 P_{i}}=\frac{270 K}{T_{f}}$

$\begin{array}{c}{T_{f}=1.9 \times 270 \mathrm{K}} \\{T_{f}=513 \mathrm{K}}\end{array}$

Hence, the heat absorbed at a constant volume V will be,

$Q=m c_{v} \Delta T$

$Q=m\left(c_{p}-R\right)\left(T_{f}-T_{i}\right)$

$Q=29 \times(24-8.31)(513-270)$

$Q=110.56743 \mathrm{kJ}$