Question

1.0-mol sample of an ideal gas is kept at 0.0°C during an expansion from 3.0 L to 10.0 L.(A) How much work is done on the gas during the expansion?

Answer 1

To calculate the work done on the gas during the expansion, you can use the formula:

$W=-P\Delta V$

where:

• $W$ represents the work done on the gas,
• $P$ is the pressure of the gas, and
• $\Delta V$ is the change in volume of the gas.

In this case, since the gas is an ideal gas, we can use the ideal gas law to determine the pressure. The ideal gas law states:

$PV=nRT$

where:

• $P$ is the pressure,
• $V$ is the volume,
• $n$ is the number of moles of gas,
• $R$ is the ideal gas constant (0.0821 L·atm/(mol·K)), and
• $T$ is the temperature in Kelvin.

Given that the temperature is 0.0°C, we need to convert it to Kelvin:

$T=0.0°C+273.15=273.15K$

Now, we can calculate the initial pressure of the gas using the ideal gas law. Since the number of moles is 1.0 mol and the volume is 3.0 L, we have:

P_1 = frac{{nRT}}{{V}} = frac{{(1.0 , ext{mol})(0.0821 , ext{L} cdot ext{atm/(mol} cdot ext{K)}})(273.15 , ext{K})}}{{3.0 , ext{L}}}

Simplifying the expression, we get:

$P_1\approx 7.16\text{atm}$

Next, we calculate the final pressure of the gas using the same formula, but with the final volume of 10.0 L:

P_2 = frac{{(1.0 , ext{mol})(0.0821 , ext{L} cdot ext{atm/(mol} cdot ext{K)}})(273.15 , ext{K})}}{{10.0 , ext{L}}}

Simplifying the expression, we get:

$P_2\approx 2.46\text{atm}$

Now that we have the initial and final pressures, as well as the change in volume, we can calculate the work done on the gas: