Question

at what temperature will a fixed amount of gas with a volume of 175 L at 15 degrees celsius and 760mmHg occupy a volume of 198L at a pressure of 640mmHg?

Answer 1

Answer: The temperature when the volume and pressure has changed is 274 K

Explanation:

To calculate the pressure when temperature and volume has changed, we use the equation given by combined gas law. The equation follows:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1,V_1\text{ and }T_1$ are the initial pressure, volume and temperature of the gas

$P_2,V_2\text{ and }T_2$ are the final pressure, volume and temperature of the gas

We are given:

$P_1=760mmHg\\V_1=175L\\T_1=15^oC=[15+273]K=288K\\P_2=640mmHg\\V_2=198L\\T_2=?K$

Putting values in above equation, we get:

$\frac{760mmHg\times 175L}{288K}=\frac{640mmHg\times 198L}{T_2}\\\\T_2=274K$

Hence, the temperature when the volume and pressure has changed is 274 K

Answer 2

Answer:

T₂ = 274.5 K = 1.35 °C

Explanation:

Given: V₁ = 175 L, P₁ = 760 mmHg, T₁ = 15°C = 288.15 K (∵ 1°C=273.15 K)

V₂ = 198 L, P₂ = 640 mmHg, T₂ = ? K

To calculate T₂, we use the General Gas Equation: $\frac{P_{1}V_{1}}{T_{1}}= \frac{P_{2}V_{2}}{T_{2}}$

$\frac{(760 mmHg)\times (175 L)}{288.15 K}= \frac{(640 mmHg)\times(198 L)}{T_{2}}$

$T_{2}= \frac{288.15 K \times 640 mmHg \times 198 L}{760 mmHg\times 175 L}$

$T_{2}= \frac{36514368}{133000} = 274.5 K$

Therefore, T₂ = 274.5 K = 1.35 °C